Solution of combined tasks. Types of tasks Task on the mixture in the problem of the problem are words: "mix", "technical", "admixture", the names of minerals or alloys. Calculation of the mass proportion of acetic acid in a mixture with formic to neutralization 7.6 g of the mixture

  1. Determine the mass Mg 3 N 2, which has completely subjected to water decomposition, if for salt formation with hydrolysis products, it was required 150 ml of a 4% hydrochloric acid solution with a density of 1.02 g / ml.
    Show

    1) Mg 3 N 2 + 6H 2 O → 3MG (OH) 2 + 2NH 3
    2) MG (OH) 2 + 2HCl → MgCl 2 + 2H 2 O
    3) NH 3 + HCl → NH 4 Cl
    N (HCl) \u003d 150 * 1.02 * 0.04 / 36.5 \u003d 0,168 mol
    Let the reaction of x mol Mg 3 N 2 be reacted. By equation 1, 3x mole Mg (OH) 2 and 2x mol NH 3 was formed. On the neutralization of 3x mol Mg (OH) 2, the 6x mole of HCl (by equation 2) was required, and the 2x mol of NH 3 was required 2 mol HCl (by equation 3), only 8 mol HCl.
    8 x \u003d 0, 168 mol,
    X \u003d 0, 021 mol,
    n (Mg 3 n 2) \u003d 0, 021 mol,
    M (Mg 3 n 2) \u003d m * n \u003d 100 * 0.021 \u003d 2.1 g
    Answer: 2.1 g
  2. Determine the mass fraction of sodium carbonate in a solution obtained by boiling 150 g of an 8.4% sodium bicarbonate solution. What volume of the 15.6% barium chloride solution (1.11 g / ml density) will react with the resulting sodium carbonate? Water evaporation can be neglected.
    Show

    1) 2NAHCO 3 - T → Na 2 CO 3 + H 2 O + CO 2
    2) Na 2 CO 3 + BaCl 2 → Baco 3 + 2 NaCl
    N (NaHCO 3) \u003d 150 * 0.084 / 84 \u003d 0.15 mol
    From equation (1) N (NaHCO 3): N (Na 2 CO 3) \u003d 2: 1 \u003d\u003e n (Na 2 CO 3) \u003d 0.075 mol.
    M (Na 2 CO 3) \u003d 0.075 ∙ 106 \u003d 7.95
    N (CO 2) \u003d 0.075 mol, M (CO 2) \u003d 0.075 ∙ 44 \u003d 3.3 g
    M (solution) \u003d 150 - 3,3 \u003d 146.7 g
    ω (Na 2 CO 3) \u003d M (Na 2 CO 3) / m (solution) \u003d 7.95 / 146.7 \u003d 0.0542 or 5.42%
    From equation (2) n (Na 2 CO 3): n (BaCl 2) \u003d 1: 1 \u003d\u003e n (BaCl 2) \u003d 0.075 mol.
    M (BaCl 2) \u003d n * m \u003d 0.075 * 208 \u003d 15.6 g
    m (solution) \u003d M (BaCl 2) / Ω \u003d 15.6 / 0.156 \u003d 100 g
    V (solution) \u003d M (solution) / ρ \u003d 100 / 1,11 \u003d 90.1 ml.
    Answer: 5.42%, 90.1 ml.
  3. In which mass ratios should be mixed with 10% solutions of sodium hydroxide and sulfuric acid to obtain a neutral sodium sulfate solution? What is the mass fraction of salt in such a solution?
    Show

    2NAOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O
    Let the mass of the NaOH solution 100 g, M (NaOH) \u003d 10 g,
    n (NaOH) \u003d 0.25 mol
    n (H 2 SO 4) \u003d 0,125 mol
    M (H 2 SO 4) \u003d 12.25 g,
    M (solution H 2 SO 4) \u003d 122.5 g
    Treason M (NaOH solution): M (solution H 2 SO 4) \u003d 1: 1,2
    n (Na 2 SO 4) \u003d 0.125 mol
    M (Na 2 SO 4) \u003d 17.75 g,
    m (solution) \u003d 100 + 122.5 g \u003d 222.5 g,
    W (Na 2 SO 4) \u003d 7.98%
  4. How many liters of chlorine (N.U.) will be released if it is to 200 ml of 35% hydrochloric acid (density of 1.17 g / ml) add during heating 26.1 g of manganese oxide (IV)? How many grams of sodium hydroxide in the cold solution reacts with this amount of chlorine?
    Show

    1) MNO 2 + 4 HCl → MnCl 2 + 2 H 2 O + CL 2
    2) 2NAOH + Cl 2 → NaCl + NaClo + H 2 O
    N (HCl) \u003d 200 * 1,17 * 0.35 / 36.5 \u003d 2.24 mol - in excess
    n (MnO 2) \u003d 26.1 / 87 \u003d 0.3 mol - in a shortcoming
    By equation (1) n (Cl 2) \u003d 0.3 mol
    V (Cl 2) \u003d 6.72 l
    By equation (2) n (NaOH) \u003d 0.6 mol, M (NaOH) \u003d 24 g
    Answer: 6.72 l, 24 g
  5. In which volume of water should be dissolved 11.2 liters of sulfur oxide (IV) (n. Y.) To obtain a solution of sulfuric acid with a mass fraction of 1%? What color will the Lacmus acquire when adding it to the resulting solution?
  6. What weight of lithium hydride should be dissolved in 100 ml of water to obtain a solution with a mass fraction of 5% hydroxide? What color will the Lacmus acquire when adding it to the resulting solution?
    Show

    Lih + H 2 O → Lioh + H 2
    Let M (Lih) \u003d x r, then m (lioh) \u003d x * 24/8 \u003d 3x
    m (p-ra) \u003d m (H 2 O) + m (Lih) - M (H 2)
    M (p-ra) \u003d x + 100 - x / 4 \u003d 0.75x + 100
    W \u003d M (V-BA) * 100% / M (RR)
    3x / (0.75x + 100) \u003d 0.05
    3x \u003d 0.038x + 5
    2.96x \u003d 5.
    x \u003d 1.7 g
  7. In which mass of the solution with the mass fraction of Na 2 SO 4 10%, it is necessary to dissolve 200 g Na 2 SO 4 × 10H 2 o to obtain a solution with a mass fraction of sodium sulfate 16%? What environment will have the resulting solution?
    Show

    Let the mass of the solution - X, it contains 0.1x g Na 2 SO 4.
    200 g of crystallohydrate was added, in which sodium sulfate mass 200 * 142/322 \u003d 88.2 g.
    (0.1x +88.2) / (x + 200) \u003d 0.16
    0.1x +88.2 \u003d 0.16x + 32
    0.06x \u003d 56,2
    x \u003d 937.
    Answer: 937 g, neutral.
  8. Ammonia gaseous ammonia, which was highlighted at a boiling of 160 g of a 7% solution of potassium hydroxide with 9.0 g of ammonium chloride, was dissolved in 75 g of water. Determine the mass fraction of ammonia in the resulting solution.
  9. Ammonia, elected at a boiling of 80 g of a 14% solution of potassium hydroxide with 8.03 g of ammonium chloride, was dissolved in water. Calculate how many milliliters 5% nitric acid A density of 1.02 g / ml will go to the neutralization of the resulting ammonia solution.
    Show

    KOH + NH 4 CL → NH 3 + H 2 O + KCL
    NH 3 + HNO 3 → NH 4 NO 3
    n (NH 4 Cl) \u003d 8.03 / 53,5 \u003d 0.15 mol
    M (KOH) \u003d 80 * 0.14 \u003d 11.2 g
    N (KOH) \u003d 11.2 / 56 \u003d 0.2 mol
    Con in excess.
    Further calculations are under deficiency
    n (NH 4 Cl) \u003d n (NH 3) \u003d 0.15 mol
    n (hno 3) \u003d 0.15 mol
    M (HNO 3) \u003d N * m \u003d 0.15 * 63 \u003d 9.45 g

    M (P-RR HNO 3) \u003d M (V-BA) * 100% / w \u003d 9.45 / 0.05 \u003d 189 g
    m \u003d v * ρ
    V \u003d m / ρ \u003d 189 / 1.02 \u003d 185.3 ml
    Answer: 185.3 ml
  10. Calcium carbide is treated with excess water. The separated gas took the volume of 4.48 l (N.U.). Calculate what volume of 20% hydrochloric acid with a density of 1.10 g / ml will go to the complete neutralization of the alkali formed from calcium carbide.
    Show

    1) CAC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
    n (C 2 H 2) \u003d V / V m \u003d 4.48 / 22.4 \u003d 0.2 mol
    From equation (1) \u003d\u003e n (C 2 H 2) \u003d N (Ca (OH) 2)
    N (Ca (OH) 2) \u003d 0.2 mol
    2) CA (OH) 2 + 2HCl \u003d CaCl 2 + 2H 2 O
    From equation (2) \u003d\u003e n (Ca (OH) 2): N (HCl) \u003d 1: 2 \u003d\u003e n (HCl) \u003d 0.4 mol
    M (HCl) \u003d n * m \u003d 0.4 * 36,5 \u003d 14.6g
    M (HCl solution) \u003d 14.6 / 0.2 \u003d 73g
    V (HCl solution) \u003d 73 / 1,1 \u003d 66.4 ml
  11. Calculate how the volume of the 10% solution of chloride chloride with a density of 1.05 g / ml will go to the complete neutralization of calcium hydroxide formed in the hydrolysis of calcium carbide, if the gas distinguished during hydrolysis took the volume of 8.96 l (N.U.).
    Show

    CAC 2 + 2 H 2 O → Ca (OH) 2 + C 2 H 2
    Ca (OH) 2 + 2 HCl → CaCl 2 + 2 H 2 O
    n (C 2 H 2) \u003d V / V N.U. \u003d 8.96 / 22.4 \u003d 0.4 mol
    n (C 2 H 2) \u003d N Ca (OH) 2 \u003d 0.4 mol
    N (HCl) \u003d 0.4 * 2 \u003d 0.8 mol
    M (V-va HCl) \u003d n * m \u003d 0.8 * 36,5 \u003d 29.2 g
    W \u003d M (V-BA) * 100% / M (RR)
    M (p-ra) \u003d m (V-BA) * 100% / w \u003d 29.2 / 0.1 \u003d 292 g
    m \u003d v * ρ
    V \u003d m / ρ \u003d 292 / 1.05 \u003d 278ml
    Answer: 278 ml
  12. Aluminum carbide was treated 200 g of 30% sulfuric acid solution. At the same time, methane took the volume of 4.48 l (N.U.). Calculate the mass fraction of sulfuric acid in the resulting solution.
    Show

    Al 4 C 3 + 6H 2 SO 4 → 2AL 2 (SO 4) 3 + 3CH 4
    N (CH 4) \u003d V / VM \u003d 4,48 / 22,4 \u003d 0.2 mol
    M (CH 4) \u003d M * m \u003d 0.2 * 16 \u003d 3.2 g
    M (Al 4 C 3) \u003d 1/3 * 0.2 * 144 \u003d 9.6 g
    Entered into the reaction, according to the equation N (H 2 SO 4) \u003d 0.4 mol,
    M (H 2 SO 4) \u003d 0.4 * 98 \u003d 39.2 g.
    Initially added M (H 2 SO 4) \u003d m (p-ra) * ω \u003d 200 g * 0.3 \u003d 60 gosta m (H 2 SO 4) \u003d 60 - 39.2 \u003d 20.8 g
    M (H 2 SO 4) \u003d 0.21 * 98 \u003d 20.8 g
    M (P-PA) \u003d M (Al 4 C 3) + M (P-PA H 2 SO 4) - M (CH 4)
    M (P-PA) \u003d 9.6 g + 200 g - 3.2 g \u003d 206.4 g
    ω (H 2 SO 4) \u003d M (H 2 SO 4) / m (p-ra) \u003d 20.8 / 206.4 * 100% \u003d 10%
  13. When processing aluminum carbide with a solution of hydrochloric acid, the mass of which is 320 g and the mass fraction of HCl 22%, 6.72 l (N.O.) methane was released. Calculate the mass fraction of hydrochloric acid in the resulting solution.
    Show

    Al 4 C 3 + 12HCl → 4AlCl 3 + 3CH 4 N (CH 4) \u003d V / VM \u003d b, 72 / 22.4 \u003d 0.3 mol;
    By equation N (HCl) \u003d 4 N (CH 4) \u003d 1.2 mol
    M (HCl) \u003d m (p-ra) * ω \u003d 320 · 0.22 \u003d 70.4 g;
    Entered into the reaction M (HCl) \u003d 1.2 · 36.5 \u003d 43.8 g.
    It remains M (HCl) \u003d 70.4 - 43.8 \u003d 26.6 g.
    M (P-PA) \u003d 320 g + m (Al 4 C 3) - M (CH 4),
    According to the equation n (Al 4 C 3) \u003d 1/3 n (CH 4) \u003d 0.1 mol;
    M (Al 4 C 3) \u003d 0.1 · 144 \u003d 14.4 g,
    M (CH 4) \u003d 0.3 · 16 \u003d 4.8 g,
    M (P-PA) \u003d 320 g + 14.4 g - 4.8 g \u003d 329.6 g
    Ω (HCl) \u003d 26.6 / 329.6 · 100% \u003d 8.07%
  14. Calcium hydride was introduced into an excess solution of hydrochloric acid (mass of an acid solution 150 g, mass fraction of HCl 20%). At the same time, 6.72 liters (N.O.) of hydrogen were separated. Calculate the mass fraction of calcium chloride in the resulting solution.
    Show

    n (CaCl 2) \u003d ½ n (H 2) \u003d 0.15 mol
    M (CaCl 2) \u003d 111 * 0.15 \u003d 16.65 g
    W (CaCl 2) \u003d M V-V / M Rr \u003d 16.65 / 155.7 \u003d 0.1069 or 10.69%
    Answer: w (CaCl 2) \u003d 10.69%
  15. 125 ml of a 5% lithium hydroxide solution was mixed (R \u003d 1.05 g / ml) and 100 ml of a 5% nitric acid solution (ρ \u003d 1.03 g / ml). Determine the medium of the resulting solution and the mass fraction of lithium nitrate in it.
    Show

    Lioh + HNO₃ \u003d LINO₃ + H₂O
    M (LioH solution) \u003d V × ρ \u003d 125 ml × 1.05 g / ml \u003d 131.25 g
    M (Lioh) \u003d 131.25 g × 0.05 \u003d 6.563 g
    n (lioh) \u003d m / m \u003d 6,563 / 24 \u003d 0,273 mol
    M (HNO₃ :) solution \u003d V × ρ \u003d 100 ml × 1.03 g / ml \u003d 103 g
    M (hno₃) \u003d 103 g × 0.05 \u003d 5.15 g
    n (hno₃) \u003d 5.15 / 63 \u003d 0,0817 mole
    Lioh Dan in excess, calculation of acid.
    n (Lino₃) \u003d 0,0817 mol
    M (Lino₃) \u003d n × m \u003d 0.0817 × 69 \u003d 5.64 g
    M (the resulting solution) \u003d m (p-ra lioh) + m (p-ra hno₃) \u003d 131.25 g + 103 g \u003d 234.25 g
    Ω (LINO₃) \u003d 5.64 / 234.25 × 100% \u003d 2.4%
    Answer: Alkaline, 2.4%;
  16. Phosphorus oxide (V) weighing 1.42 g was dissolved in 60 g of 8.2% orthophosphoric acid and the resulting solution was boiled. Which salt and in what quantity is formed if adding 3.92 g of potassium hydroxide add to the resulting solution?
  17. Sulfur oxide (VI) weighing 8 g was dissolved in 110 g of 8% sulfuric acid. Which salt and in what amount is formed if you add 10.6 g of potassium hydroxide to the resulting solution?
    Show

    SO 3 + H 2 O \u003d H 2 SO 4
    n \u003d m / mm (SO 3) \u003d 80 g / mol,
    n (SO 3) \u003d 8/80 \u003d 0.1 mol.
    By equation (1) n (H 2 SO 4) \u003d n (SO 3) \u003d 0.1 mol,
    N (KOH) \u003d 10.6 / 56 \u003d 0.19 mol.
    In the initial solution n (H 2 SO 4) \u003d 110 * 0.08 / 98 \u003d 0, 09 mol.
    After adding sulfur oxide n (H 2 SO 4) \u003d 0.09 + 0.1 \u003d 0.19 mol.
    The amounts of alkali and acid substance are correlated as 1: 1, it means the acidic salt is formed
    H 2 SO 4 + KOH \u003d KHSO 4 + H 2 O
    n (H 2 SO 4) \u003d n (koh) \u003d n (khso 4) \u003d 0.19 mol
    Answer: KHSO 4, 0.19 mol.
  18. Ammonia, which was distinguished by the interaction of 107 g of a 20% solution of ammonium chloride with 150 g of a 18% sodium hydroxide solution, completely reacted with 60% orthophosphoric acid with the formation of ammonium dihydrophosphate. Determine the mass fraction of sodium chloride in solution and the necessary mass of 60% phosphoric acid solution.
    Show

    NH 4 Cl + NaOH \u003d NaCl + NH 3 + H 2 O
    M (NH 4 Cl) \u003d 107 g ∙ 0.2 \u003d 21.4 g
    n (NH 4 Cl) \u003d 21.4 g / 53.5 g / mol \u003d 0.4 mol
    M (NaOH) \u003d 150 g ∙ 0.18 \u003d 27 g
    N (NaOH) \u003d 27 g / 40 g / mol \u003d 0.675 mol, therefore, NaOH in excess
    N (NaCl) \u003d n (NH4Cl) \u003d 0.4 mol
    M (NaCl) \u003d 0.4 ∙ 58.5 \u003d 23.4 g
    N (NH 3) \u003d N (NH 4 Cl) \u003d 0.4 mol
    M (NH 3) \u003d 0.4 ∙ 17 \u003d 6.8 g
    m (p-ra) \u003d m (p-rane 4 cl) + m (p-raneoh) - M (NH 3) \u003d 107 + 150 - 6.8 \u003d 250.2 g
    W (NaCl) \u003d 23.4 / 250.2 \u003d 0.094 or 9.4%
    NH 3 + H 3 PO 4 \u003d NH 4 H 2 PO 4
    N (NH 3) \u003d n (H 3 PO 4) \u003d 0.4 mol
    M (H 3 PO 4) \u003d 98 ∙ 0.4 \u003d 39.2 g
    M (p-ra H 3 PO 4) \u003d 39.2 / 0.6 \u003d 65.3 g
  19. The hydrogen sulfide distinguished by the interaction of an excess of concentrated sulfuric acid with 1.44 g of magnesium, passed through 160 g of 1.5% bromine solution. Determine the mass of the precipitate and the mass fraction of acid in the resulting solution.
    Show

    4mg + 5h 2 SO 4 \u003d 4mgSO 4 + H 2 S + 4H 2 O
    H 2 S + BR 2 \u003d 2HBR + S ↓
    n (Mg) \u003d m / m \u003d 1.44 g: 24 g / mol \u003d 0.06 mol
    n (H 2 S) \u003d ¼ n (Mg) \u003d 0.015 mol
    M (H 2 S) \u003d n * m \u003d 0.015 mol * 34 g / mol \u003d 0.51 mol
    M (V-VA BR 2) \u003d 160 g * 0.015 \u003d 2.4 g
    n (br 2) \u003d m / m \u003d 2.4 g: 160 g / mol \u003d 0,015 mol
    N (HBr) \u003d 2N (Br 2) \u003d 0.03 mol
    M (HBr) \u003d n * m \u003d 0.03 mol * 81 g / mol \u003d 2.43 g
    n (s) \u003d n (br 2) \u003d 0,015 mol
    M (s) \u003d n * m \u003d 0.015 mol * 32 g / mol \u003d 0.48 g
    m (p-ra) \u003d m (H 2 S) + m (p-ra Br 2) -m (s) \u003d 0.51 g + 160 g - 0.48 \u003d 160.03 g
    W (HBr) \u003d M (HBr) / m (p-ra) \u003d 2.43 g / 160.03 g \u003d 0.015 or 1.5%
    Answer: M (S) \u003d 0, 48 g, W (HBr) \u003d 1.5%

  20. Chlorine without a residue reacted with 228.58 ml of a 5% solution of Naon (density of 1.05 g / ml) at elevated temperature. Determine the composition of the resulting solution and calculate the mass fractions of substances in this solution.
    Show

    6NAOH + 3CL 2 \u003d 5NACL + NACLO 3 + 3H 2 O (at T)
    m solution \u003d 228.58 ∙ 1.05 \u003d 240g;
    M (NaOH) \u003d 240 ∙ 0.05 \u003d 12g.
    n (NaOH) \u003d 12/40 \u003d 0.3 mol;
    n (Cl 2) \u003d 0.15 mol;
    n (naCl) \u003d 0.25 mol;
    n (naclo 3) \u003d 0.05 mol
    M (NaCl) \u003d 58.5 ∙ 0.25 \u003d 14,625g;
    M (Naclo 3) \u003d 106.5 ∙ 0.05 \u003d 5,325 g:
    m solution \u003d 240 + m (Cl 2) \u003d 240 + 71 ∙ 0.15 \u003d 240 + 10,65 \u003d 250,65
    W (NaCl) \u003d 14,625 / 250.65 \u003d 0.0583 or 5.83%
    W (NaClo 3) \u003d 5,325 / 250.65 \u003d 0.0212 or 2.12%
  21. Copper weighing 6.4 g was treated with 100 ml of 30% nitric acid (ρ \u003d 1,153 g / ml). For complete binding of products, a 200 g of sodium hydroxide solution was added to the resulting solution. Determine the mass fraction of alkalis in the resulting solution.
    Show

    3CU + 8HNO 3 \u003d 3CU (NO 3) 2 + 2NO + 4H 2 O
    M (HNO 3) \u003d 100 ∙ 0.3 ∙ 1,153 \u003d 34.59
    n (hno 3) \u003d 34,59 / 63 \u003d 0.55 mol, n (Cu) \u003d 6.4 / 64 \u003d 0.1 mol
    n (hno 3) gave \u003d 0.55 - 8/3 ∙ 0.1 \u003d 0.28 mol
    Cu (NO 3) 2 + 2 NaOH \u003d Cu (OH) 2 + 2 Nano 3
    HNO 3 + NaOH \u003d Nano 3 + H 2 O
    N (NaOH) \u003d n (HNO 3) AS + 2N (Cu (NO 3) 2) \u003d 0.28 + 0.1 ∙ 2 \u003d 0.48 mol
    M (NaOH) \u003d 0.48 ∙ 40 \u003d 19.2g.
    W (NaOH) \u003d 19.2 / 200 \u003d 0.096 or 9.6%
  22. In 60 g of 18% orthophosphoric acid, 2.84 g of phosphorus oxide (V) was dissolved and the resulting solution was boiled. What salt and in what amount is formed if you add 30 g sodium hydroxide to the resulting solution?
    Show

    1) 3H 2 O + P 2 O 5 → 2H 3 PO 4
    In the initial solution M (H 3 PO 4) \u003d M (solution) * ω \u003d 60 * 0.18 \u003d 10.8 g
    N (P 2 O 5) \u003d m / m \u003d 2.84 / 142 \u003d 0.02 mole
    As a result of the reaction, M (H 3 PO 4) \u003d 0.04 * 98 \u003d 3.92 g
    Total M (H 3 PO 4) \u003d 3.92 + 10.8 \u003d 14.72
    N (H 3 PO 4) \u003d m / m \u003d 14,72 / 98 \u003d 0.15 mol
    n (NaOH) \u003d m / m \u003d 30/40 \u003d 0.75 mol - in excess, salt is average.
    2) 3NAOH + H 3 PO 4 → Na 3 PO 4 + 3H 2 O
    By equation (2) n (Na 3 PO 4) \u003d n (H 3 PO 4) \u003d 0.15 mol
    M (Na 3 PO 4) \u003d 0.15 * 164 \u003d 24.6g
    Answer: 24.6 g
  23. Ammonia with a volume of 4.48 l (N.U.) was missed by 200 g of 4.9% of the solution of orthophosphoric acid. Name the salt resulting from the reaction and determine its mass.
  24. 5.6 l (N.O.) hydrogen sulfide reacted without a residue with 59.02 ml of a 20% solution of Koh (density of 1.186 g / ml). Determine the mass of salts obtained as a result of this chemical reaction.
    Show

    m (p-ra KOH) \u003d V * ρ \u003d 1.186 * 59.02 \u003d 70g,
    m (koh) \u003d m (p-rakoh) * ω \u003d 70g * 0.2 \u003d 14g,
    n (koh) \u003d m / m \u003d 14/56 \u003d 0.25 mol,
    n (H 2 S) \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol.
    The amount of hydrogen sulfide is equal to the amount of alkali, therefore, an acidic salt is formed - hydrosulfide by reaction: H 2 S + KOH \u003d KNS + H 2 O
    By equation N (KHS) \u003d 0.25 mol,
    M (KHS) \u003d M * n \u003d 72 · 0.25 \u003d 18
    Answer: 18 g

  25. The neutralization of 7.6 g of a mixture of formic and acetic acids was spent by 35 ml of a 20% solution of potassium hydroxide (density of 1.20 g / ml). Calculate the mass of acetic acid and its mass fraction in the initial mixture of acids.
    Show

    Coxy + kon \u003d nsokov + H 2 o
    CH 3 coam + Kon \u003d CH 3 COI + H 2 O
    M (P-PAKOH) \u003d V (P-PA) * ρ \u003d 35 * 1.2 \u003d 42
    M (KOH) \u003d M (P-PA) * Ω (KOH) \u003d 42 * 0.2 \u003d 8.4 g
    N (KOH) \u003d M (KOH) / M (KOH) \u003d 8.4 / 56 \u003d 0.15 mol
    Let N (NSonon) \u003d X mol, and n (CH 3 of the coxy) \u003d at mole.
    M (HCOOH) \u003d N (HCOOH) * M (HCOOH) \u003d x * 46 g
    M (CH 3 COOH) \u003d N (CH 3 COOH) * M (CH 3 COOH) \u003d y * 60 g
    Make a system of equations:
    x + y \u003d 0.15
    60u + 46x \u003d 7.6
    I solve the system: x \u003d 0.1 mol, y \u003d 0.05 mol
    M (CH 3 COOH) \u003d N (CH 3 COOH) * M (CH 3 COOH) \u003d 0.05 * 60 \u003d 3 g
    ω (CH 3 COOH) \u003d M (CH 3 COOH) / M (mixtures) \u003d 3 / 7.6 \u003d 0.395 or 39.5%.
    Answer: 39.5%
  26. 100 ml of 30% hydrochloric acid solution was mixed (R \u003d 1.11 g / ml) and 300 ml of a 20% sodium hydroxide solution (R \u003d 1.10 g / ml). How many milliliters of water should be added to the mixture obtained so that the mass fraction of sodium perchlorate in it would be 8%?
    Show

    HCLO 4 + NaOH \u003d Naclo 4 + H 2 O
    M (PR NaOH) \u003d V * ρ \u003d 300 * 1,10 \u003d 330
    N (NaOH) \u003d M (PR NaOH) * ω / m \u003d 330 * 0.2 / 40 \u003d 1.65 mol - in excess.
    m (p-p HCLO 4) \u003d V * ρ \u003d 100 * 1,11 \u003d 111
    n (HCLO 4) \u003d 111 * 0.3 / 100.5 \u003d 0.331 mol,
    By equation N (HCLO 4) \u003d N (NaClo 4) \u003d 0.331 mol,
    M (NaClo 4) \u003d n * m \u003d 0.331 * 122.5 \u003d 40.5 g
    Let the mass of the added water - x
    40.5 / (111 + 330 + x) \u003d 0.08
    where x \u003d 65.3 g g g
    V (H 2 O) \u003d 65.3 ml.
    Answer: 65.3 ml
  27. 6.4 g of calcium carbide was added to 100 ml of a 5% solution of hydrochloric acid (1.02 g / ml density). How many milliliters of 15% nitric acid (1.08 g / ml density) should be added to the mixture obtained for its complete neutralization?
    Show

    1) CAC 2 + 2HCl \u003d CaCl 2 + C 2 H 2
    2) CAC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
    3) CA (OH) 2 + 2HNO 3 \u003d Ca (NO 3) 2 + 2H 2 O
    N (HCl) \u003d M (HCl) / M (HCl) \u003d Vr-ra (HCl) * (HCl) * P (HCl) / M (HCl) \u003d 100 * 0.05 * 1.02 / 36.5 \u003d 0.14 mol,
    n (Cac 2) \u003d m / m \u003d 6.4 / 64 \u003d 0.1 mol.
    According to equation (1) N (CAC 2): N (HCl) \u003d 1: 2 \u003d\u003e CAC 2 - in excess.
    Entered into the reaction N (CAC 2) \u003d N (HCl) / 2 \u003d 0.07 mol.
    N (CAC 2) \u003d 0.1 - 0.07 \u003d 0.03 mol.
    According to equation (2) N (CAC 2) \u003d N (Ca (OH) 2) \u003d 0.03 mol.
    By equation (3) n (Ca (OH) 2): n (hnO3) \u003d 1: 2 \u003d\u003e
    N (HNO 3) \u003d 2N (Ca (OH) 2) \u003d 0.06 mol.
    VR-RA (HNO 3) \u003d M (P-RG) / ρ
    m (p-ra) \u003d m (hno 3) / ω \u003d 0.06 * 63 / 0.15 \u003d 25.2 g,
    V (p-rahno 3) \u003d 25.2 / 1.08 \u003d 23.3 ml.
    Answer: 23.3 ml
  28. Sodium nitrite weighing 13.8 g was made when heated in 220 g of ammonium chloride solution with a mass fraction of 10%. What volume (N.U.) of nitrogen is distinguished and what is the mass fraction of ammonium chloride in the resulting solution?
    Show

    Nano 2 + NH 4 CL \u003d N 2 + NaCl + 2H 2 O
    n (Nano 2) \u003d 13.8 / 69 \u003d 0.2 mol
    n (NH 4 Cl) \u003d 220 · 0.1 / 53,5 \u003d 0.41 mol
    NH 4 Cl - in excess (N 2) \u003d N (Nano 2) \u003d 0.2 mol
    V (n 2) \u003d 0.2 mol · 22.4 l / mol \u003d 4.48 l
    Calculate the mass of ammonium chloride remaining in excess:
    N (NH 4 CL) gave \u003d 0.41 - 0.2 \u003d 0.21 mol
    M (NH 4 CL) Large \u003d 0.21 · 53,5 \u003d 11.2 Grascuting the mass fraction of ammonium chloride:
    M (P-PA) \u003d 13.8 + 220 - 0.2 · 28 \u003d 228.2 g
    Ω (NH 4 CL) \u003d 11.2 / 228.2 \u003d 0.049 or 4.9% Answer:
    V (N 2) \u003d 4.48 l
    Ω (NH 4 Cl) \u003d 4.9%
  29. Potassium nitrite weighing 8.5 g was made when heated in 270 g of ammonium bromide solution with a mass fraction of 12%. What volume (n.у) nitrogen is distinguished and what is the mass fraction of ammonium bromide in the resulting solution?
    Show

    KNO 2 + NH 4 Br \u003d N 2 + KBR + 2H 2 O
    M (V-va NH 4 br) \u003d 270 g * 0.12 \u003d 32.4 g
    n (NH 4 br) \u003d m / m \u003d 32.4 g: 98 g / mol \u003d 0.33 mol
    n (KNO 2) \u003d m / m \u003d 8.5 g: 85 g / mol \u003d 0.1 mol
    N (NH 4), prowork. with KNO 2 \u003d 0.33 mol - 0.1 mol \u003d 0.23 mol (t. n (KNO 2): N (NH 4 br) \u003d 1: 1)
    M (NH 4 BR) remaining in the final solution \u003d n * m \u003d 0.23 mol * 98 g / mol \u003d 22.54 g
    n (n 2) \u003d n (KNO 2) \u003d 0, 1 mol
    (N 2) \u003d n * m \u003d 0.1 mol * 28 g / mol \u003d 2.8 g
    V (n 2) \u003d n * m \u003d 0.1 mol * 22.4 l / mol \u003d 2.24 l
    m (finite. P-ra) \u003d M (KNO 2) + m (p-ra NH 4 br) - m (n 2) \u003d 8.5 g + 270 g - 2.8 g \u003d 275, 7 g
    W (NH 4 BR in fin. P-RE) \u003d 22.54 g: 275.7 g \u003d 8%
    Answer: V (n 2) \u003d 2.24 l; W (NH 4 br) \u003d 8%
  30. 300 ml of sulfuric acid solution with a mass fraction of 10% (density of 1.05 g / ml) and 200 ml of potassium hydroxide solution with a mass fraction of 20% were mixed with a mass fraction of 20% (1.10 g / ml density). How many milliliters of water should be added to the mixture obtained so that the mass proportion of salt in it was 7%?
  31. In 120 ml of a nitric acid solution with a mass fraction of 7% (a density of 1.03 g / ml), 12.8 g of calcium carbide was made. How many milliliters of 20% hydrochloric acid (1.10 g / ml density) should be added to the mixture obtained for its complete neutralization?
    Show

    1) CAC 2 + 2HNO 3 \u003d Ca (NO 3) 2 + C 2 H 2.
    N (CAC 2) \u003d 12.8 / 64 \u003d 0.2 mol
    n (hno 3) \u003d (0.07 · 1.03 · 120) / 63 \u003d 0,137 mol
    CAC 2 - excess.2) CAC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2.
    n (Ca (OH) 2) \u003d 0.2 - 0.137 / 2 \u003d 0.13 mol3) Ca (OH) 2 + 2HCl \u003d CaCl 2 + 2H 2 O.
    N (HCl) \u003d 0.13 · 2 \u003d 0.26 mol
    m (p-ra) \u003d m (nsl) / w \u003d (0.26 · 36,5) / 0.2 \u003d 47.45 g
    V (p-ra HCl) \u003d m (p-ra) / ρ \u003d 47.45 / 1,10 \u003d 43.1 ml.
    Answer: 43.1 ml.
  32. To the solution obtained by adding 4 g of potassium hydride to 100 ml of water, 100 ml of 39% nitric acid solution was adhered (R \u003d 1.24 g / ml). Determine the mass fractions of all substances (including water) in the final solution.
    Show

    KH + H 2 O \u003d KOH + H 2
    N (KH) \u003d 4 g: 40 g / mol \u003d 0.1 mol
    n (H 2 O) \u003d 100 g: 18 g / mol \u003d 5.6 mol
    KH - in a shortage, then n (koh) \u003d n (kh) \u003d 0.1 MolkOH + H 2 NO 3 \u003d KNO 3 + H 2 O
    a) M (KOH) \u003d 0.1 * 56 \u003d 5.6
    B) M V-BA (HNO 2) \u003d 100 ml * 1.24 g / ml * 0.39 \u003d 48.36 g
    n (hno 3) \u003d 48.36 g: 63 g / mol \u003d 0.77 mol
    HNO 3 in excess, N Bulk. (HNO 3) \u003d 0.77 - 0.1 \u003d 0.67 mol
    M (HNO 3) \u003d 0.67 * 63 \u003d 42.21 g
    M (p-ra) \u003d 4g + 100g + 124g - 0.2g \u003d 227.8
    3) W (KNO 3) \u003d M (KNO 3): M (P-PA) \u003d (0.1 mol * 101 g / mol) 227.8 g * 100% \u003d 4.4%
    W (HNO 3) \u003d 42,21: 227.8 * 100% \u003d 18.5
    W (H 2 O) \u003d 100% - (W (KNO 3) + W (HNO 3)) \u003d 77.1%
    Answer:
    W (KNO 3) \u003d 4.4%
    W (hno 3) \u003d 18.5%
    W (H 2 O) \u003d 77.1%

    1) 2NA 2 O 2 + 2H 2 O \u003d 4NAOH + O 2
    2) 2NAOH + H 2 SO 4 \u003d Na 2 SO 4 + 2H 2 O
    M (solution H 2 SO 4) \u003d 300 * 1.08 \u003d 324 g
    M (H 2 SO 4) \u003d 0.1 * 324 \u003d 32.4g
    N (H 2 SO 4) \u003d 32.4 / 98 \u003d 0.33 mol
    N (NaOH): n (H 2 SO 4) \u003d 2: 1 \u003d\u003e n (NaOH) \u003d 0.33 * 2 \u003d 0.66 mol
    n (Na 2 O 2): n (NaOH) \u003d 1: 2 \u003d\u003e n (Na 2 O 2) \u003d 0.66 / 2 \u003d 0.33 mol
    M (Na 2 O 2) \u003d n * m \u003d 0.33 * 78 \u003d 25.7 g
    n (Na 2 O 2): n (O 2) \u003d 2: 1 \u003d\u003e n (O 2) \u003d 0.33 / 2 \u003d 0,165 mol
    V (O 2) \u003d 0.165 * 22.4 \u003d 3.7 liters
    Answer: M (Na 2 O 2) \u003d 25.7 g; V (O 2) \u003d 3.7 liters

  33. When heated, potassium bicarbonate turns into carbonate. Calculate the mass fraction of potassium bicarbonate in the source solution, the heating of which is the 8% potassium carbonate solution.
  34. When interacting in the sulfurish medium, 17.4 g of manganese dioxide with 58 g of potassium bromide at 77% output was separated by bromine. What volume (n.у) is proposed to be involved with the resulting bromine?
  35. Carbon dioxide with a volume of 5.6 liters (N.U.) was passed through 164 ml of a 20% sodium hydroxide solution (ρ \u003d 1.22 g / ml). Determine the composition and mass fractions of substances in the resulting solution.
  36. In a 15% sulfuric acid solution weighing 300 g soluted aluminum carbide. At the same time, methane occupied the volume of 2.24 l (N.U.). Calculate the mass fraction of sulfuric acid in the resulting solution.
  37. In excess oxygen burned 8 g of sulfur. The resulting gas was missed by 200 g of an 8% sodium hydroxide solution. Determine the mass fractions of salts in the resulting solution.
  38. A mixture of aluminum and iron sawdust was treated with an excess of dilute hydrochloric acid, while 8.96 liters (N.O.) of hydrogen were separated. If the same mass of the mixture is treated with an excess solution of sodium hydroxide, then 6.72 liters (N.O.) of hydrogen are separated. Calculate the mass fraction of iron in the source mixture.
  39. The mixture of magnesium and zinc sawdust was treated with an excess of diluted sulfuric acid, while 22.4 liters (N.O.) of hydrogen were separated. If the same mass of the mixture is treated with an excess solution of sodium hydroxide, then 13.44 liters (N.O.) of hydrogen are separated. Calculate the mass fraction of magnesium in the initial mixture.
    2khco 3 + H 2 SO 4 \u003d K 2 SO 4 + 2CO 2 + 2H 2 O (2)

    Na 2 CO 3 + 2HCl \u003d NaCl + CO 2 + H 2 O
    NaHCO 3 + HCl \u003d NaCl + CO 2 + H 2 O
    NaHCO 3 + NaOH \u003d Na 2 CO 3 + H 2 O
    1) findquantity NaHCO 3.
    n (NaOH) \u003d (NaHCO 3) \u003d M (NaOH). Ω (NaOH) / M (NaOH) \u003d 80 * 0.1 / 40 \u003d 0.2 mol
    2) findmass NaHCO 3.
    m (NaHCO 3) \u003d N (NaHCO 3) * M (NaHCO 3) \u003d 0.2. 84 \u003d 16.8 g
    3) we find the amount of Na 2 CO 3
    n (HCl) \u003d M (HCl). Ω (HCl) / M (HCl) \u003d 73 * 0.2 / 36.5 \u003d 0.4mol
    n (Na 2 CO 3) \u003d (n (HCl) - n (NaOH)) / 2 \u003d 0.1 mol
    4) we find a massNa. 2 Co. 3
    m (Na 2 CO 3) \u003d n. M \u003d 0.1. 106 \u003d 10.6 g
    5) we find a massive dassuNa 2 CO 3
    Ω (Na 2 CO 3) \u003d M (Na 2 CO 3) / M (mixture). 100% \u003d 38.7%
    M (mixture). \u003d M (Na 2 CO 3) + M (NaHCO 3) \u003d 27.4 g
    Answer: Ω (Na 2 CO 3) \u003d 38.7%

    1) 2NA + 2H 2 O → 2NAOH + H 2
    2) Na 2 O + H 2 O → 2naoh
    1) we find the numberNa.
    n (H 2) \u003d V / VM \u003d 4.48 / 22.4 \u003d 0.2 mol
    By equation (1) n (Na) \u003d N 1 (NaOh) \u003d 2. N (H 2) \u003d 0.4 mol
    2) We will find a massNa.
    M (Na) \u003d n. M \u003d 0.4. 23 \u003d 9.2 g
    3)     we find the numberNaoh.
    n (NaOH) \u003d M (P-P). Ω / M (NaOH) \u003d 240. 0.1 / 40 \u003d 0.6 mol
    4) we find the numberNa. 2 O.
    n (Na 2 O) \u003d (n (NaOH) - N 1 (NaOh)) / 2 \u003d 0.1 mol
    M (Na 2 O) \u003d n. M \u003d 0.1. 62 \u003d 6.2g
    5) findm.assova Dolia.Na.
    Ω (Na) \u003d M (Na) / M (mixtures). 100% \u003d 59.7%
    m (mixtures) \u003d m (na) + m (na 2 O) \u003d 15.4
    Answer: Ω (Na) \u003d59,7%

    H 2 SO 4 + Na 2 CO 3 → Na 2 SO 4 + CO 2 + H 2
    1) Find a total numberH 2 SO 4
    n 1 (H 2 SO 4) \u003d m. ω / m \u003d 490. 0.4 / 98 \u003d 2 mole
    M (H 2 SO 4) \u003d 2 + 32 + 64 \u003d 98 g / mol
    2) FindH 2 SO 4rearachingfrom Na 2 CO 3
    n (Na 2 CO 3) \u003d n (Na 2 CO 3. 10H 2 O) \u003d M / M \u003d 143/286 \u003d 0.5 mol
    M (Na 2 CO 3. 10H 2 O) \u003d (46 + 12 + 48) + (10. 18) \u003d 286 g / mol
    By equation N (H 2 SO 4) \u003d N (Na 2 CO 3) \u003d 0,5mol
    3) findH. 2 SO. 4 Rearamed S.Naoh.
    n (H 2 SO 4) \u003d N 1 -N \u003d 2-0.5 \u003d 1.5mol
    4) Find a lot of NaOH
    n (NaOH) \u003d 2N (H 2 SO 4) \u003d 2. 1,5 \u003d 3 mole
    M (NaOH) \u003d n. M \u003d 3. 40 \u003d 120 g
    M (NaOH) \u003d 23 + 16 + 1 \u003d 18 g / mol
    5) we find a mass shareNaoh.
    Ω (NaOH) \u003d m (ML) /. M (P-P) * 100% \u003d 120/1200. 100% \u003d 10%
    Answer: M (NaOH) \u003d 120g; Ω (NaOH) \u003d 10%

    48) A mixture of magnesium and aluminum sawdust was treated with an excess of diluted hydrochloric acid, while 11.2 liters (N.O.) of hydrogen were separated. If the same mass of the mixture is treated with an excess solution of potassium hydroxide, then 6.72 liters (N.O.) of hydrogen are separated. Calculate the mass fraction of magnesium in the initial mixture.
    49) Calcium carbide weighing 6.4 g was dissolved in 87 ml of bromomic acid (ρ \u003d 1.12 g / ml) with a mass fraction of 20%. What is the mass proportion of bromomomodor in the resulting solution?

In this problem, there are two parallel reactions involving a mixture of substances.

Task 3.9.
On the neutralization of 7.6 g of a mixture of formic and acetic acids, 35 ml of a 20% solution of potassium hydroxide (p \u003d 1.2 g / ml) was consumed. Calculate the mass of acetic acid and its mass fraction in the initial mixture of acids.
Given:
mass of the mixture of acids: M (mixtures of acids) \u003d 7.6 g;
the volume of the solution of Kon: V p-ra (con) \u003d 35 ml;
mass fraction of con in the initial R-RE: (CO) in Ex. p-re \u003d \u200b\u200b20%;
the density of the initial solution Kon: P Ex. p-ra (con) \u003d 1.2 g / ml.
To find: Acetic acid mass: M (CH 3 coxy);
mass fraction CH 3 coxy in the original mixture: (CH 3 of the coxy) \u003d?
Decision:

In the interaction of a mixture of acids with potassium hydroxide, two reactions are at the same time:

CH 3 coxy + Kon → CH 3 COI + H 2 O
Nson + Kon → Nsokok + N 2

The solution of the problem is possible on a conventional algorithm with the compilation of the mathematical equation. The solution scheme can be represented as follows:

1. Denote a mass of acetic acid in the initial mixture of "A":

m (CH 3 coxy) \u003d A G.

Then the mass of formic acid can be determined by difference:

m (Nsonon) \u003d M (acid mixtures) - M (CH 3 coxy) \u003d (7.6 - a)

2. According to the equation of acetic acid reaction with a solution of alkali, we will determine the mass of the con which was consumed in this interaction.

3. Similarly, according to the reaction equation with formic acid, we find a mass of con which was spent in the second interaction.

4. We find the total mass of the con, which was spent in two reactions:

m (con) \u003d m (con) in the R-│ with UKS K-TA + M (CO) in the R-│ with Moore K-TU \u003d
\u003d 0.933. a + (9.25 - 1,217. A) \u003d (9.25 - 0.284. A).

5. We find a mass of the cross, which was contained in 200 ml of the initial solution:

6. We equate the expression obtained in 4 actions to the value of the mass of CO from 5:

9.25 - 0.284. A \u003d 8.4.

Received a mathematical equation. Solving it, we find the value of the magnitude of "A":

The magnitude of "A" we denoted a mass of acetic acid:

m (CH 3 coxy) \u003d 3 g

7. We find a mass fraction of CH 3 coxy in the source mixture of acids

Answer: M (CH 3 COOH) \u003d 3 g; (CH 3 COOH) \u003d 39.5%.

Types of tasks Task on the mixture in the condition of the problem are words: "mix", "technical", "impressory", the names of minerals or alloys of the problem of solutions in the condition of the problem are words: "solution", "mass fraction of a dissolved substance" problem for excess The disadvantage in the condition of the task is information about both reagents of the task for the product output in the problem of the problem there are words: "The output of the substance", the "mass fraction of the product output" of the tasks in which products of one reaction are used to carry out another reaction.






Here is the problem ... for the reaction of 6.3 g of aluminum and magnesium mixture with sulfuric acid, 275.8 ml of a 10% sulfuric acid solution (density is 1.066 g / ml) is required. Determine what the mass of a 20% solution of barium chloride will be required for the total precipitation of metals sulfates from the resulting solution.




We define the amount of solid sulfuric acid in the solution: m (p-ra) \u003d V (R-ra). \u003d 278.8. 1.066 \u003d 294 (g) M (H 2 SO 4) \u003d M (R-RA). W (H 2 SO 4) \u003d, 1 \u003d 29.4g N (H 2 SO 4) \u003d 29.4 / 98 \u003d 0.3 mol 1 more score!




We define a mass of solution: n (bacl2) \u003d 0.3 mol M (BaCl2) \u003d, 3 \u003d 62.4 (g) m (p-ra BaCl2) \u003d 62.4: 0.2 \u003d 312 (d) answer: M (P-ra BaCl 2) \u003d 312 g and another 1 point total - 4 points total - 4 points












Elements of problem solving and evaluation of results 1. The equations of all reactions are properly written correctly. 2. Found "Number of substances" (volume, mass) of the starting materials 3. Composed system of equations for finding the amount of substances in the mixture. 4. Calculate the end data, which is asked in the task. All items are correctly completed - 4 points All error is allowed - 3 points are allowed two errors - 2 points are made three errors - 1 point All items are made incorrect - 0 points


The tasks for self-solving a mixture of zinc and zinc carbonate were treated with an excessive solution of hydrochloric acid, and 13.44l gas was separated (H.). The gas was burned, combustion products were cooled to the same temperature, while the gas volume decreased to 8.96l. What composition had the initial mixture of substances?




Homework 1 Sodium nitrite weighing 13.8 g was introduced when heated in 220 g of ammonium chloride solution with a mass fraction of 10%. What volume (n.у) nitrogen is highlighted and what is the mass fraction of ammonium chloride in the resulting solution? Answer: w (NH 4 Cl) \u003d 4.9%


The homework 2 of the potassium nitrite weighing 8.5 g was made when heated in 270 g of ammonium bromide solution with a mass fraction of 12%. What volume (n.у) nitrogen is distinguished and what is the mass fraction of ammonium bromide in the resulting solution? Answer: V (n 2) \u003d 2.24 l, w (NH 4 br) \u003d 8.2%


Homework 3 was mixed with 300 ml of sulfuric acid solution with a mass fraction of 10% (density of 1.05 g / ml) and 200 ml of potassium hydroxide solution with a mass fraction of 20% (1.10 g / ml density). What volume of water should be added to the mixture obtained, so that the mass proportion of salt in it was 7%? Answer: V \u003d 262.9 l


The homework 4 in 120 ml of a solution of nitric acid with a mass fraction of 7% (a density of 1.03 g / ml) was made by 12.8 g of calcium carbide. What volume of 20% hydrochloric acid (1.10 g / ml density) should be added to the resulting mixture for its complete neutralization? Answer: V \u003d 43.1 ml


Homework 5 When interacting in the sulfur-acid medium, 8.7 g of manganese dioxide with 22.4 g of potassium bromide was separated by bromine, which was 88% practical. What volume (n.) ethylene can react with the resulting bromine? Answer: V \u003d 1, 86 l



  1. Determine the mass Mg 3 N 2, which has completely subjected to water decomposition, if for salt formation with hydrolysis products, it was required 150 ml of a 4% hydrochloric acid solution with a density of 1.02 g / ml.
    Show

    1) Mg 3 N 2 + 6H 2 O → 3MG (OH) 2 + 2NH 3
    2) MG (OH) 2 + 2HCl → MgCl 2 + 2H 2 O
    3) NH 3 + HCl → NH 4 Cl
    N (HCl) \u003d 150 * 1.02 * 0.04 / 36.5 \u003d 0,168 mol
    Let the reaction of x mol Mg 3 N 2 be reacted. By equation 1, 3x mole Mg (OH) 2 and 2x mol NH 3 was formed. On the neutralization of 3x mol Mg (OH) 2, the 6x mole of HCl (by equation 2) was required, and the 2x mol of NH 3 was required 2 mol HCl (by equation 3), only 8 mol HCl.
    8 x \u003d 0, 168 mol,
    X \u003d 0, 021 mol,
    n (Mg 3 n 2) \u003d 0, 021 mol,
    M (Mg 3 n 2) \u003d m * n \u003d 100 * 0.021 \u003d 2.1 g
    Answer: 2.1 g
  2. Determine the mass fraction of sodium carbonate in a solution obtained by boiling 150 g of an 8.4% sodium bicarbonate solution. What volume of the 15.6% barium chloride solution (1.11 g / ml density) will react with the resulting sodium carbonate? Water evaporation can be neglected.
    Show

    1) 2NAHCO 3 - T → Na 2 CO 3 + H 2 O + CO 2
    2) Na 2 CO 3 + BaCl 2 → Baco 3 + 2 NaCl
    N (NaHCO 3) \u003d 150 * 0.084 / 84 \u003d 0.15 mol
    From equation (1) N (NaHCO 3): N (Na 2 CO 3) \u003d 2: 1 \u003d\u003e n (Na 2 CO 3) \u003d 0.075 mol.
    M (Na 2 CO 3) \u003d 0.075 ∙ 106 \u003d 7.95
    N (CO 2) \u003d 0.075 mol, M (CO 2) \u003d 0.075 ∙ 44 \u003d 3.3 g
    M (solution) \u003d 150 - 3,3 \u003d 146.7 g
    ω (Na 2 CO 3) \u003d M (Na 2 CO 3) / m (solution) \u003d 7.95 / 146.7 \u003d 0.0542 or 5.42%
    From equation (2) n (Na 2 CO 3): n (BaCl 2) \u003d 1: 1 \u003d\u003e n (BaCl 2) \u003d 0.075 mol.
    M (BaCl 2) \u003d n * m \u003d 0.075 * 208 \u003d 15.6 g
    m (solution) \u003d M (BaCl 2) / Ω \u003d 15.6 / 0.156 \u003d 100 g
    V (solution) \u003d M (solution) / ρ \u003d 100 / 1,11 \u003d 90.1 ml.
    Answer: 5.42%, 90.1 ml.
  3. In which mass ratios should be mixed with 10% solutions of sodium hydroxide and sulfuric acid to obtain a neutral sodium sulfate solution? What is the mass fraction of salt in such a solution?
    Show

    2NAOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O
    Let the mass of the NaOH solution 100 g, M (NaOH) \u003d 10 g,
    n (NaOH) \u003d 0.25 mol
    n (H 2 SO 4) \u003d 0,125 mol
    M (H 2 SO 4) \u003d 12.25 g,
    M (solution H 2 SO 4) \u003d 122.5 g
    Treason M (NaOH solution): M (solution H 2 SO 4) \u003d 1: 1,2
    n (Na 2 SO 4) \u003d 0.125 mol
    M (Na 2 SO 4) \u003d 17.75 g,
    m (solution) \u003d 100 + 122.5 g \u003d 222.5 g,
    W (Na 2 SO 4) \u003d 7.98%
  4. How many liters of chlorine (N.U.) will be released if it is to 200 ml of 35% hydrochloric acid (density of 1.17 g / ml) add during heating 26.1 g of manganese oxide (IV)? How many grams of sodium hydroxide in the cold solution reacts with this amount of chlorine?
    Show

    1) MNO 2 + 4 HCl → MnCl 2 + 2 H 2 O + CL 2
    2) 2NAOH + Cl 2 → NaCl + NaClo + H 2 O
    N (HCl) \u003d 200 * 1,17 * 0.35 / 36.5 \u003d 2.24 mol - in excess
    n (MnO 2) \u003d 26.1 / 87 \u003d 0.3 mol - in a shortcoming
    By equation (1) n (Cl 2) \u003d 0.3 mol
    V (Cl 2) \u003d 6.72 l
    By equation (2) n (NaOH) \u003d 0.6 mol, M (NaOH) \u003d 24 g
    Answer: 6.72 l, 24 g
  5. In which volume of water should be dissolved 11.2 liters of sulfur oxide (IV) (n. Y.) To obtain a solution of sulfuric acid with a mass fraction of 1%? What color will the Lacmus acquire when adding it to the resulting solution?
  6. What weight of lithium hydride should be dissolved in 100 ml of water to obtain a solution with a mass fraction of 5% hydroxide? What color will the Lacmus acquire when adding it to the resulting solution?
    Show

    Lih + H 2 O → Lioh + H 2
    Let M (Lih) \u003d x r, then m (lioh) \u003d x * 24/8 \u003d 3x
    m (p-ra) \u003d m (H 2 O) + m (Lih) - M (H 2)
    M (p-ra) \u003d x + 100 - x / 4 \u003d 0.75x + 100
    W \u003d M (V-BA) * 100% / M (RR)
    3x / (0.75x + 100) \u003d 0.05
    3x \u003d 0.038x + 5
    2.96x \u003d 5.
    x \u003d 1.7 g
  7. In which mass of the solution with the mass fraction of Na 2 SO 4 10%, it is necessary to dissolve 200 g Na 2 SO 4 × 10H 2 o to obtain a solution with a mass fraction of sodium sulfate 16%? What environment will have the resulting solution?
    Show

    Let the mass of the solution - X, it contains 0.1x g Na 2 SO 4.
    200 g of crystallohydrate was added, in which sodium sulfate mass 200 * 142/322 \u003d 88.2 g.
    (0.1x +88.2) / (x + 200) \u003d 0.16
    0.1x +88.2 \u003d 0.16x + 32
    0.06x \u003d 56,2
    x \u003d 937.
    Answer: 937 g, neutral.
  8. Ammonia gaseous ammonia, which was highlighted at a boiling of 160 g of a 7% solution of potassium hydroxide with 9.0 g of ammonium chloride, was dissolved in 75 g of water. Determine the mass fraction of ammonia in the resulting solution.
  9. Ammonia, elected at a boiling of 80 g of a 14% solution of potassium hydroxide with 8.03 g of ammonium chloride, was dissolved in water. Calculate how many milliliters of 5% nitric acid with a density of 1.02 g / ml will go to the neutralization of the obtained immaterial solution.
    Show

    KOH + NH 4 CL → NH 3 + H 2 O + KCL
    NH 3 + HNO 3 → NH 4 NO 3
    n (NH 4 Cl) \u003d 8.03 / 53,5 \u003d 0.15 mol
    M (KOH) \u003d 80 * 0.14 \u003d 11.2 g
    N (KOH) \u003d 11.2 / 56 \u003d 0.2 mol
    Con in excess.
    Further calculations are under deficiency
    n (NH 4 Cl) \u003d n (NH 3) \u003d 0.15 mol
    n (hno 3) \u003d 0.15 mol
    M (HNO 3) \u003d N * m \u003d 0.15 * 63 \u003d 9.45 g

    M (P-RR HNO 3) \u003d M (V-BA) * 100% / w \u003d 9.45 / 0.05 \u003d 189 g
    m \u003d v * ρ
    V \u003d m / ρ \u003d 189 / 1.02 \u003d 185.3 ml
    Answer: 185.3 ml
  10. Calcium carbide is treated with excess water. The separated gas took the volume of 4.48 l (N.U.). Calculate what volume of 20% hydrochloric acid with a density of 1.10 g / ml will go to the complete neutralization of the alkali formed from calcium carbide.
    Show

    1) CAC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
    n (C 2 H 2) \u003d V / V m \u003d 4.48 / 22.4 \u003d 0.2 mol
    From equation (1) \u003d\u003e n (C 2 H 2) \u003d N (Ca (OH) 2)
    N (Ca (OH) 2) \u003d 0.2 mol
    2) CA (OH) 2 + 2HCl \u003d CaCl 2 + 2H 2 O
    From equation (2) \u003d\u003e n (Ca (OH) 2): N (HCl) \u003d 1: 2 \u003d\u003e n (HCl) \u003d 0.4 mol
    M (HCl) \u003d n * m \u003d 0.4 * 36,5 \u003d 14.6g
    M (HCl solution) \u003d 14.6 / 0.2 \u003d 73g
    V (HCl solution) \u003d 73 / 1,1 \u003d 66.4 ml
  11. Calculate how the volume of the 10% solution of chloride chloride with a density of 1.05 g / ml will go to the complete neutralization of calcium hydroxide formed in the hydrolysis of calcium carbide, if the gas distinguished during hydrolysis took the volume of 8.96 l (N.U.).
    Show

    CAC 2 + 2 H 2 O → Ca (OH) 2 + C 2 H 2
    Ca (OH) 2 + 2 HCl → CaCl 2 + 2 H 2 O
    n (C 2 H 2) \u003d V / V N.U. \u003d 8.96 / 22.4 \u003d 0.4 mol
    n (C 2 H 2) \u003d N Ca (OH) 2 \u003d 0.4 mol
    N (HCl) \u003d 0.4 * 2 \u003d 0.8 mol
    M (V-va HCl) \u003d n * m \u003d 0.8 * 36,5 \u003d 29.2 g
    W \u003d M (V-BA) * 100% / M (RR)
    M (p-ra) \u003d m (V-BA) * 100% / w \u003d 29.2 / 0.1 \u003d 292 g
    m \u003d v * ρ
    V \u003d m / ρ \u003d 292 / 1.05 \u003d 278ml
    Answer: 278 ml
  12. Aluminum carbide was treated 200 g of 30% sulfuric acid solution. At the same time, methane took the volume of 4.48 l (N.U.). Calculate the mass fraction of sulfuric acid in the resulting solution.
    Show

    Al 4 C 3 + 6H 2 SO 4 → 2AL 2 (SO 4) 3 + 3CH 4
    N (CH 4) \u003d V / VM \u003d 4,48 / 22,4 \u003d 0.2 mol
    M (CH 4) \u003d M * m \u003d 0.2 * 16 \u003d 3.2 g
    M (Al 4 C 3) \u003d 1/3 * 0.2 * 144 \u003d 9.6 g
    Entered into the reaction, according to the equation N (H 2 SO 4) \u003d 0.4 mol,
    M (H 2 SO 4) \u003d 0.4 * 98 \u003d 39.2 g.
    Initially added M (H 2 SO 4) \u003d m (p-ra) * ω \u003d 200 g * 0.3 \u003d 60 gosta m (H 2 SO 4) \u003d 60 - 39.2 \u003d 20.8 g
    M (H 2 SO 4) \u003d 0.21 * 98 \u003d 20.8 g
    M (P-PA) \u003d M (Al 4 C 3) + M (P-PA H 2 SO 4) - M (CH 4)
    M (P-PA) \u003d 9.6 g + 200 g - 3.2 g \u003d 206.4 g
    ω (H 2 SO 4) \u003d M (H 2 SO 4) / m (p-ra) \u003d 20.8 / 206.4 * 100% \u003d 10%
  13. When processing aluminum carbide with a solution of hydrochloric acid, the mass of which is 320 g and the mass fraction of HCl 22%, 6.72 l (N.O.) methane was released. Calculate the mass fraction of hydrochloric acid in the resulting solution.
    Show

    Al 4 C 3 + 12HCl → 4AlCl 3 + 3CH 4 N (CH 4) \u003d V / VM \u003d b, 72 / 22.4 \u003d 0.3 mol;
    By equation N (HCl) \u003d 4 N (CH 4) \u003d 1.2 mol
    M (HCl) \u003d m (p-ra) * ω \u003d 320 · 0.22 \u003d 70.4 g;
    Entered into the reaction M (HCl) \u003d 1.2 · 36.5 \u003d 43.8 g.
    It remains M (HCl) \u003d 70.4 - 43.8 \u003d 26.6 g.
    M (P-PA) \u003d 320 g + m (Al 4 C 3) - M (CH 4),
    According to the equation n (Al 4 C 3) \u003d 1/3 n (CH 4) \u003d 0.1 mol;
    M (Al 4 C 3) \u003d 0.1 · 144 \u003d 14.4 g,
    M (CH 4) \u003d 0.3 · 16 \u003d 4.8 g,
    M (P-PA) \u003d 320 g + 14.4 g - 4.8 g \u003d 329.6 g
    Ω (HCl) \u003d 26.6 / 329.6 · 100% \u003d 8.07%
  14. Calcium hydride was introduced into an excess solution of hydrochloric acid (mass of an acid solution 150 g, mass fraction of HCl 20%). At the same time, 6.72 liters (N.O.) of hydrogen were separated. Calculate the mass fraction of calcium chloride in the resulting solution.
    Show

    n (CaCl 2) \u003d ½ n (H 2) \u003d 0.15 mol
    M (CaCl 2) \u003d 111 * 0.15 \u003d 16.65 g
    W (CaCl 2) \u003d M V-V / M Rr \u003d 16.65 / 155.7 \u003d 0.1069 or 10.69%
    Answer: w (CaCl 2) \u003d 10.69%
  15. 125 ml of a 5% lithium hydroxide solution was mixed (R \u003d 1.05 g / ml) and 100 ml of a 5% nitric acid solution (ρ \u003d 1.03 g / ml). Determine the medium of the resulting solution and the mass fraction of lithium nitrate in it.
    Show

    Lioh + HNO₃ \u003d LINO₃ + H₂O
    M (LioH solution) \u003d V × ρ \u003d 125 ml × 1.05 g / ml \u003d 131.25 g
    M (Lioh) \u003d 131.25 g × 0.05 \u003d 6.563 g
    n (lioh) \u003d m / m \u003d 6,563 / 24 \u003d 0,273 mol
    M (HNO₃ :) solution \u003d V × ρ \u003d 100 ml × 1.03 g / ml \u003d 103 g
    M (hno₃) \u003d 103 g × 0.05 \u003d 5.15 g
    n (hno₃) \u003d 5.15 / 63 \u003d 0,0817 mole
    Lioh Dan in excess, calculation of acid.
    n (Lino₃) \u003d 0,0817 mol
    M (Lino₃) \u003d n × m \u003d 0.0817 × 69 \u003d 5.64 g
    M (the resulting solution) \u003d m (p-ra lioh) + m (p-ra hno₃) \u003d 131.25 g + 103 g \u003d 234.25 g
    Ω (LINO₃) \u003d 5.64 / 234.25 × 100% \u003d 2.4%
    Answer: Alkaline, 2.4%;
  16. Phosphorus oxide (V) weighing 1.42 g was dissolved in 60 g of 8.2% orthophosphoric acid and the resulting solution was boiled. Which salt and in what quantity is formed if adding 3.92 g of potassium hydroxide add to the resulting solution?
  17. Sulfur oxide (VI) weighing 8 g was dissolved in 110 g of 8% sulfuric acid. Which salt and in what amount is formed if you add 10.6 g of potassium hydroxide to the resulting solution?
    Show

    SO 3 + H 2 O \u003d H 2 SO 4
    n \u003d m / mm (SO 3) \u003d 80 g / mol,
    n (SO 3) \u003d 8/80 \u003d 0.1 mol.
    By equation (1) n (H 2 SO 4) \u003d n (SO 3) \u003d 0.1 mol,
    N (KOH) \u003d 10.6 / 56 \u003d 0.19 mol.
    In the initial solution n (H 2 SO 4) \u003d 110 * 0.08 / 98 \u003d 0, 09 mol.
    After adding sulfur oxide n (H 2 SO 4) \u003d 0.09 + 0.1 \u003d 0.19 mol.
    The amounts of alkali and acid substance are correlated as 1: 1, it means the acidic salt is formed
    H 2 SO 4 + KOH \u003d KHSO 4 + H 2 O
    n (H 2 SO 4) \u003d n (koh) \u003d n (khso 4) \u003d 0.19 mol
    Answer: KHSO 4, 0.19 mol.
  18. Ammonia, which was distinguished by the interaction of 107 g of a 20% solution of ammonium chloride with 150 g of a 18% sodium hydroxide solution, completely reacted with 60% orthophosphoric acid with the formation of ammonium dihydrophosphate. Determine the mass fraction of sodium chloride in solution and the necessary mass of 60% phosphoric acid solution.
    Show

    NH 4 Cl + NaOH \u003d NaCl + NH 3 + H 2 O
    M (NH 4 Cl) \u003d 107 g ∙ 0.2 \u003d 21.4 g
    n (NH 4 Cl) \u003d 21.4 g / 53.5 g / mol \u003d 0.4 mol
    M (NaOH) \u003d 150 g ∙ 0.18 \u003d 27 g
    N (NaOH) \u003d 27 g / 40 g / mol \u003d 0.675 mol, therefore, NaOH in excess
    N (NaCl) \u003d n (NH4Cl) \u003d 0.4 mol
    M (NaCl) \u003d 0.4 ∙ 58.5 \u003d 23.4 g
    N (NH 3) \u003d N (NH 4 Cl) \u003d 0.4 mol
    M (NH 3) \u003d 0.4 ∙ 17 \u003d 6.8 g
    m (p-ra) \u003d m (p-rane 4 cl) + m (p-raneoh) - M (NH 3) \u003d 107 + 150 - 6.8 \u003d 250.2 g
    W (NaCl) \u003d 23.4 / 250.2 \u003d 0.094 or 9.4%
    NH 3 + H 3 PO 4 \u003d NH 4 H 2 PO 4
    N (NH 3) \u003d n (H 3 PO 4) \u003d 0.4 mol
    M (H 3 PO 4) \u003d 98 ∙ 0.4 \u003d 39.2 g
    M (p-ra H 3 PO 4) \u003d 39.2 / 0.6 \u003d 65.3 g
  19. The hydrogen sulfide distinguished by the interaction of an excess of concentrated sulfuric acid with 1.44 g of magnesium, passed through 160 g of 1.5% bromine solution. Determine the mass of the precipitate and the mass fraction of acid in the resulting solution.
    Show

    4mg + 5h 2 SO 4 \u003d 4mgSO 4 + H 2 S + 4H 2 O
    H 2 S + BR 2 \u003d 2HBR + S ↓
    n (Mg) \u003d m / m \u003d 1.44 g: 24 g / mol \u003d 0.06 mol
    n (H 2 S) \u003d ¼ n (Mg) \u003d 0.015 mol
    M (H 2 S) \u003d n * m \u003d 0.015 mol * 34 g / mol \u003d 0.51 mol
    M (V-VA BR 2) \u003d 160 g * 0.015 \u003d 2.4 g
    n (br 2) \u003d m / m \u003d 2.4 g: 160 g / mol \u003d 0,015 mol
    N (HBr) \u003d 2N (Br 2) \u003d 0.03 mol
    M (HBr) \u003d n * m \u003d 0.03 mol * 81 g / mol \u003d 2.43 g
    n (s) \u003d n (br 2) \u003d 0,015 mol
    M (s) \u003d n * m \u003d 0.015 mol * 32 g / mol \u003d 0.48 g
    m (p-ra) \u003d m (H 2 S) + m (p-ra Br 2) -m (s) \u003d 0.51 g + 160 g - 0.48 \u003d 160.03 g
    W (HBr) \u003d M (HBr) / m (p-ra) \u003d 2.43 g / 160.03 g \u003d 0.015 or 1.5%
    Answer: M (S) \u003d 0, 48 g, W (HBr) \u003d 1.5%

  20. Chlorine without a residue reacted with 228.58 ml of a 5% solution of Naon (density of 1.05 g / ml) at elevated temperature. Determine the composition of the resulting solution and calculate the mass fractions of substances in this solution.
    Show

    6NAOH + 3CL 2 \u003d 5NACL + NACLO 3 + 3H 2 O (at T)
    m solution \u003d 228.58 ∙ 1.05 \u003d 240g;
    M (NaOH) \u003d 240 ∙ 0.05 \u003d 12g.
    n (NaOH) \u003d 12/40 \u003d 0.3 mol;
    n (Cl 2) \u003d 0.15 mol;
    n (naCl) \u003d 0.25 mol;
    n (naclo 3) \u003d 0.05 mol
    M (NaCl) \u003d 58.5 ∙ 0.25 \u003d 14,625g;
    M (Naclo 3) \u003d 106.5 ∙ 0.05 \u003d 5,325 g:
    m solution \u003d 240 + m (Cl 2) \u003d 240 + 71 ∙ 0.15 \u003d 240 + 10,65 \u003d 250,65
    W (NaCl) \u003d 14,625 / 250.65 \u003d 0.0583 or 5.83%
    W (NaClo 3) \u003d 5,325 / 250.65 \u003d 0.0212 or 2.12%
  21. Copper weighing 6.4 g was treated with 100 ml of 30% nitric acid (ρ \u003d 1,153 g / ml). For complete binding of products, a 200 g of sodium hydroxide solution was added to the resulting solution. Determine the mass fraction of alkalis in the resulting solution.
    Show

    3CU + 8HNO 3 \u003d 3CU (NO 3) 2 + 2NO + 4H 2 O
    M (HNO 3) \u003d 100 ∙ 0.3 ∙ 1,153 \u003d 34.59
    n (hno 3) \u003d 34,59 / 63 \u003d 0.55 mol, n (Cu) \u003d 6.4 / 64 \u003d 0.1 mol
    n (hno 3) gave \u003d 0.55 - 8/3 ∙ 0.1 \u003d 0.28 mol
    Cu (NO 3) 2 + 2 NaOH \u003d Cu (OH) 2 + 2 Nano 3
    HNO 3 + NaOH \u003d Nano 3 + H 2 O
    N (NaOH) \u003d n (HNO 3) AS + 2N (Cu (NO 3) 2) \u003d 0.28 + 0.1 ∙ 2 \u003d 0.48 mol
    M (NaOH) \u003d 0.48 ∙ 40 \u003d 19.2g.
    W (NaOH) \u003d 19.2 / 200 \u003d 0.096 or 9.6%
  22. In 60 g of 18% orthophosphoric acid, 2.84 g of phosphorus oxide (V) was dissolved and the resulting solution was boiled. What salt and in what amount is formed if you add 30 g sodium hydroxide to the resulting solution?
    Show

    1) 3H 2 O + P 2 O 5 → 2H 3 PO 4
    In the initial solution M (H 3 PO 4) \u003d M (solution) * ω \u003d 60 * 0.18 \u003d 10.8 g
    N (P 2 O 5) \u003d m / m \u003d 2.84 / 142 \u003d 0.02 mole
    As a result of the reaction, M (H 3 PO 4) \u003d 0.04 * 98 \u003d 3.92 g
    Total M (H 3 PO 4) \u003d 3.92 + 10.8 \u003d 14.72
    N (H 3 PO 4) \u003d m / m \u003d 14,72 / 98 \u003d 0.15 mol
    n (NaOH) \u003d m / m \u003d 30/40 \u003d 0.75 mol - in excess, salt is average.
    2) 3NAOH + H 3 PO 4 → Na 3 PO 4 + 3H 2 O
    By equation (2) n (Na 3 PO 4) \u003d n (H 3 PO 4) \u003d 0.15 mol
    M (Na 3 PO 4) \u003d 0.15 * 164 \u003d 24.6g
    Answer: 24.6 g
  23. Ammonia with a volume of 4.48 l (N.U.) was missed by 200 g of 4.9% of the solution of orthophosphoric acid. Name the salt resulting from the reaction and determine its mass.
  24. 5.6 l (N.O.) hydrogen sulfide reacted without a residue with 59.02 ml of a 20% solution of Koh (density of 1.186 g / ml). Determine the mass of salts obtained as a result of this chemical reaction.
    Show

    m (p-ra KOH) \u003d V * ρ \u003d 1.186 * 59.02 \u003d 70g,
    m (koh) \u003d m (p-rakoh) * ω \u003d 70g * 0.2 \u003d 14g,
    n (koh) \u003d m / m \u003d 14/56 \u003d 0.25 mol,
    n (H 2 S) \u003d V / Vm \u003d 5.6 / 22.4 \u003d 0.25 mol.
    The amount of hydrogen sulfide is equal to the amount of alkali, therefore, an acidic salt is formed - hydrosulfide by reaction: H 2 S + KOH \u003d KNS + H 2 O
    By equation N (KHS) \u003d 0.25 mol,
    M (KHS) \u003d M * n \u003d 72 · 0.25 \u003d 18
    Answer: 18 g

  25. The neutralization of 7.6 g of a mixture of formic and acetic acids was spent by 35 ml of a 20% solution of potassium hydroxide (density of 1.20 g / ml). Calculate the mass of acetic acid and its mass fraction in the initial mixture of acids.
    Show

    Coxy + kon \u003d nsokov + H 2 o
    CH 3 coam + Kon \u003d CH 3 COI + H 2 O
    M (P-PAKOH) \u003d V (P-PA) * ρ \u003d 35 * 1.2 \u003d 42
    M (KOH) \u003d M (P-PA) * Ω (KOH) \u003d 42 * 0.2 \u003d 8.4 g
    N (KOH) \u003d M (KOH) / M (KOH) \u003d 8.4 / 56 \u003d 0.15 mol
    Let N (NSonon) \u003d X mol, and n (CH 3 of the coxy) \u003d at mole.
    M (HCOOH) \u003d N (HCOOH) * M (HCOOH) \u003d x * 46 g
    M (CH 3 COOH) \u003d N (CH 3 COOH) * M (CH 3 COOH) \u003d y * 60 g
    Make a system of equations:
    x + y \u003d 0.15
    60u + 46x \u003d 7.6
    I solve the system: x \u003d 0.1 mol, y \u003d 0.05 mol
    M (CH 3 COOH) \u003d N (CH 3 COOH) * M (CH 3 COOH) \u003d 0.05 * 60 \u003d 3 g
    ω (CH 3 COOH) \u003d M (CH 3 COOH) / M (mixtures) \u003d 3 / 7.6 \u003d 0.395 or 39.5%.
    Answer: 39.5%
  26. 100 ml of 30% hydrochloric acid solution was mixed (R \u003d 1.11 g / ml) and 300 ml of a 20% sodium hydroxide solution (R \u003d 1.10 g / ml). How many milliliters of water should be added to the mixture obtained so that the mass fraction of sodium perchlorate in it would be 8%?
    Show

    HCLO 4 + NaOH \u003d Naclo 4 + H 2 O
    M (PR NaOH) \u003d V * ρ \u003d 300 * 1,10 \u003d 330
    N (NaOH) \u003d M (PR NaOH) * ω / m \u003d 330 * 0.2 / 40 \u003d 1.65 mol - in excess.
    m (p-p HCLO 4) \u003d V * ρ \u003d 100 * 1,11 \u003d 111
    n (HCLO 4) \u003d 111 * 0.3 / 100.5 \u003d 0.331 mol,
    By equation N (HCLO 4) \u003d N (NaClo 4) \u003d 0.331 mol,
    M (NaClo 4) \u003d n * m \u003d 0.331 * 122.5 \u003d 40.5 g
    Let the mass of the added water - x
    40.5 / (111 + 330 + x) \u003d 0.08
    where x \u003d 65.3 g g g
    V (H 2 O) \u003d 65.3 ml.
    Answer: 65.3 ml
  27. 6.4 g of calcium carbide was added to 100 ml of a 5% solution of hydrochloric acid (1.02 g / ml density). How many milliliters of 15% nitric acid (1.08 g / ml density) should be added to the mixture obtained for its complete neutralization?
    Show

    1) CAC 2 + 2HCl \u003d CaCl 2 + C 2 H 2
    2) CAC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2
    3) CA (OH) 2 + 2HNO 3 \u003d Ca (NO 3) 2 + 2H 2 O
    N (HCl) \u003d M (HCl) / M (HCl) \u003d Vr-ra (HCl) * (HCl) * P (HCl) / M (HCl) \u003d 100 * 0.05 * 1.02 / 36.5 \u003d 0.14 mol,
    n (Cac 2) \u003d m / m \u003d 6.4 / 64 \u003d 0.1 mol.
    According to equation (1) N (CAC 2): N (HCl) \u003d 1: 2 \u003d\u003e CAC 2 - in excess.
    Entered into the reaction N (CAC 2) \u003d N (HCl) / 2 \u003d 0.07 mol.
    N (CAC 2) \u003d 0.1 - 0.07 \u003d 0.03 mol.
    According to equation (2) N (CAC 2) \u003d N (Ca (OH) 2) \u003d 0.03 mol.
    By equation (3) n (Ca (OH) 2): n (hnO3) \u003d 1: 2 \u003d\u003e
    N (HNO 3) \u003d 2N (Ca (OH) 2) \u003d 0.06 mol.
    VR-RA (HNO 3) \u003d M (P-RG) / ρ
    m (p-ra) \u003d m (hno 3) / ω \u003d 0.06 * 63 / 0.15 \u003d 25.2 g,
    V (p-rahno 3) \u003d 25.2 / 1.08 \u003d 23.3 ml.
    Answer: 23.3 ml
  28. Sodium nitrite weighing 13.8 g was made when heated in 220 g of ammonium chloride solution with a mass fraction of 10%. What volume (N.U.) of nitrogen is distinguished and what is the mass fraction of ammonium chloride in the resulting solution?
    Show

    Nano 2 + NH 4 CL \u003d N 2 + NaCl + 2H 2 O
    n (Nano 2) \u003d 13.8 / 69 \u003d 0.2 mol
    n (NH 4 Cl) \u003d 220 · 0.1 / 53,5 \u003d 0.41 mol
    NH 4 Cl - in excess (N 2) \u003d N (Nano 2) \u003d 0.2 mol
    V (n 2) \u003d 0.2 mol · 22.4 l / mol \u003d 4.48 l
    Calculate the mass of ammonium chloride remaining in excess:
    N (NH 4 CL) gave \u003d 0.41 - 0.2 \u003d 0.21 mol
    M (NH 4 CL) Large \u003d 0.21 · 53,5 \u003d 11.2 Grascuting the mass fraction of ammonium chloride:
    M (P-PA) \u003d 13.8 + 220 - 0.2 · 28 \u003d 228.2 g
    Ω (NH 4 CL) \u003d 11.2 / 228.2 \u003d 0.049 or 4.9% Answer:
    V (N 2) \u003d 4.48 l
    Ω (NH 4 Cl) \u003d 4.9%
  29. Potassium nitrite weighing 8.5 g was made when heated in 270 g of ammonium bromide solution with a mass fraction of 12%. What volume (n.у) nitrogen is distinguished and what is the mass fraction of ammonium bromide in the resulting solution?
    Show

    KNO 2 + NH 4 Br \u003d N 2 + KBR + 2H 2 O
    M (V-va NH 4 br) \u003d 270 g * 0.12 \u003d 32.4 g
    n (NH 4 br) \u003d m / m \u003d 32.4 g: 98 g / mol \u003d 0.33 mol
    n (KNO 2) \u003d m / m \u003d 8.5 g: 85 g / mol \u003d 0.1 mol
    N (NH 4), prowork. with KNO 2 \u003d 0.33 mol - 0.1 mol \u003d 0.23 mol (t. n (KNO 2): N (NH 4 br) \u003d 1: 1)
    M (NH 4 BR) remaining in the final solution \u003d n * m \u003d 0.23 mol * 98 g / mol \u003d 22.54 g
    n (n 2) \u003d n (KNO 2) \u003d 0, 1 mol
    (N 2) \u003d n * m \u003d 0.1 mol * 28 g / mol \u003d 2.8 g
    V (n 2) \u003d n * m \u003d 0.1 mol * 22.4 l / mol \u003d 2.24 l
    m (finite. P-ra) \u003d M (KNO 2) + m (p-ra NH 4 br) - m (n 2) \u003d 8.5 g + 270 g - 2.8 g \u003d 275, 7 g
    W (NH 4 BR in fin. P-RE) \u003d 22.54 g: 275.7 g \u003d 8%
    Answer: V (n 2) \u003d 2.24 l; W (NH 4 br) \u003d 8%
  30. 300 ml of sulfuric acid solution with a mass fraction of 10% (density of 1.05 g / ml) and 200 ml of potassium hydroxide solution with a mass fraction of 20% were mixed with a mass fraction of 20% (1.10 g / ml density). How many milliliters of water should be added to the mixture obtained so that the mass proportion of salt in it was 7%?
  31. In 120 ml of a nitric acid solution with a mass fraction of 7% (a density of 1.03 g / ml), 12.8 g of calcium carbide was made. How many milliliters of 20% hydrochloric acid (1.10 g / ml density) should be added to the mixture obtained for its complete neutralization?
    Show

    1) CAC 2 + 2HNO 3 \u003d Ca (NO 3) 2 + C 2 H 2.
    N (CAC 2) \u003d 12.8 / 64 \u003d 0.2 mol
    n (hno 3) \u003d (0.07 · 1.03 · 120) / 63 \u003d 0,137 mol
    CAC 2 - excess.2) CAC 2 + 2H 2 O \u003d Ca (OH) 2 + C 2 H 2.
    n (Ca (OH) 2) \u003d 0.2 - 0.137 / 2 \u003d 0.13 mol3) Ca (OH) 2 + 2HCl \u003d CaCl 2 + 2H 2 O.
    N (HCl) \u003d 0.13 · 2 \u003d 0.26 mol
    m (p-ra) \u003d m (nsl) / w \u003d (0.26 · 36,5) / 0.2 \u003d 47.45 g
    V (p-ra HCl) \u003d m (p-ra) / ρ \u003d 47.45 / 1,10 \u003d 43.1 ml.
    Answer: 43.1 ml.
  32. To the solution obtained by adding 4 g of potassium hydride to 100 ml of water, 100 ml of 39% nitric acid solution was adhered (R \u003d 1.24 g / ml). Determine the mass fractions of all substances (including water) in the final solution.
    Show

    KH + H 2 O \u003d KOH + H 2
    N (KH) \u003d 4 g: 40 g / mol \u003d 0.1 mol
    n (H 2 O) \u003d 100 g: 18 g / mol \u003d 5.6 mol
    KH - in a shortage, then n (koh) \u003d n (kh) \u003d 0.1 MolkOH + H 2 NO 3 \u003d KNO 3 + H 2 O
    a) M (KOH) \u003d 0.1 * 56 \u003d 5.6
    B) M V-BA (HNO 2) \u003d 100 ml * 1.24 g / ml * 0.39 \u003d 48.36 g
    n (hno 3) \u003d 48.36 g: 63 g / mol \u003d 0.77 mol
    HNO 3 in excess, N Bulk. (HNO 3) \u003d 0.77 - 0.1 \u003d 0.67 mol
    M (HNO 3) \u003d 0.67 * 63 \u003d 42.21 g
    M (p-ra) \u003d 4g + 100g + 124g - 0.2g \u003d 227.8
    3) W (KNO 3) \u003d M (KNO 3): M (P-PA) \u003d (0.1 mol * 101 g / mol) 227.8 g * 100% \u003d 4.4%
    W (HNO 3) \u003d 42,21: 227.8 * 100% \u003d 18.5
    W (H 2 O) \u003d 100% - (W (KNO 3) + W (HNO 3)) \u003d 77.1%
    Answer:
    W (KNO 3) \u003d 4.4%
    W (hno 3) \u003d 18.5%
    W (H 2 O) \u003d 77.1%

    1) 2NA 2 O 2 + 2H 2 O \u003d 4NAOH + O 2
    2) 2NAOH + H 2 SO 4 \u003d Na 2 SO 4 + 2H 2 O
    M (solution H 2 SO 4) \u003d 300 * 1.08 \u003d 324 g
    M (H 2 SO 4) \u003d 0.1 * 324 \u003d 32.4g
    N (H 2 SO 4) \u003d 32.4 / 98 \u003d 0.33 mol
    N (NaOH): n (H 2 SO 4) \u003d 2: 1 \u003d\u003e n (NaOH) \u003d 0.33 * 2 \u003d 0.66 mol
    n (Na 2 O 2): n (NaOH) \u003d 1: 2 \u003d\u003e n (Na 2 O 2) \u003d 0.66 / 2 \u003d 0.33 mol
    M (Na 2 O 2) \u003d n * m \u003d 0.33 * 78 \u003d 25.7 g
    n (Na 2 O 2): n (O 2) \u003d 2: 1 \u003d\u003e n (O 2) \u003d 0.33 / 2 \u003d 0,165 mol
    V (O 2) \u003d 0.165 * 22.4 \u003d 3.7 liters
    Answer: M (Na 2 O 2) \u003d 25.7 g; V (O 2) \u003d 3.7 liters

  33. When heated, potassium bicarbonate turns into carbonate. Calculate the mass fraction of potassium bicarbonate in the source solution, the heating of which is the 8% potassium carbonate solution.
  34. When interacting in the sulfurish medium, 17.4 g of manganese dioxide with 58 g of potassium bromide at 77% output was separated by bromine. What volume (n.у) is proposed to be involved with the resulting bromine?
  35. Carbon dioxide with a volume of 5.6 liters (N.U.) was passed through 164 ml of a 20% sodium hydroxide solution (ρ \u003d 1.22 g / ml). Determine the composition and mass fractions of substances in the resulting solution.
  36. In a 15% sulfuric acid solution weighing 300 g soluted aluminum carbide. At the same time, methane occupied the volume of 2.24 l (N.U.). Calculate the mass fraction of sulfuric acid in the resulting solution.
  37. In excess oxygen burned 8 g of sulfur. The resulting gas was missed by 200 g of an 8% sodium hydroxide solution. Determine the mass fractions of salts in the resulting solution.
  38. A mixture of aluminum and iron sawdust was treated with an excess of dilute hydrochloric acid, while 8.96 liters (N.O.) of hydrogen were separated. If the same mass of the mixture is treated with an excess solution of sodium hydroxide, then 6.72 liters (N.O.) of hydrogen are separated. Calculate the mass fraction of iron in the source mixture.
  39. The mixture of magnesium and zinc sawdust was treated with an excess of diluted sulfuric acid, while 22.4 liters (N.O.) of hydrogen were separated. If the same mass of the mixture is treated with an excess solution of sodium hydroxide, then 13.44 liters (N.O.) of hydrogen are separated. Calculate the mass fraction of magnesium in the initial mixture.
    2khco 3 + H 2 SO 4 \u003d K 2 SO 4 + 2CO 2 + 2H 2 O (2)

    Na 2 CO 3 + 2HCl \u003d NaCl + CO 2 + H 2 O
    NaHCO 3 + HCl \u003d NaCl + CO 2 + H 2 O
    NaHCO 3 + NaOH \u003d Na 2 CO 3 + H 2 O
    1) findquantity NaHCO 3.
    n (NaOH) \u003d (NaHCO 3) \u003d M (NaOH). Ω (NaOH) / M (NaOH) \u003d 80 * 0.1 / 40 \u003d 0.2 mol
    2) findmass NaHCO 3.
    m (NaHCO 3) \u003d N (NaHCO 3) * M (NaHCO 3) \u003d 0.2. 84 \u003d 16.8 g
    3) we find the amount of Na 2 CO 3
    n (HCl) \u003d M (HCl). Ω (HCl) / M (HCl) \u003d 73 * 0.2 / 36.5 \u003d 0.4mol
    n (Na 2 CO 3) \u003d (n (HCl) - n (NaOH)) / 2 \u003d 0.1 mol
    4) we find a massNa. 2 Co. 3
    m (Na 2 CO 3) \u003d n. M \u003d 0.1. 106 \u003d 10.6 g
    5) we find a massive dassuNa 2 CO 3
    Ω (Na 2 CO 3) \u003d M (Na 2 CO 3) / M (mixture). 100% \u003d 38.7%
    M (mixture). \u003d M (Na 2 CO 3) + M (NaHCO 3) \u003d 27.4 g
    Answer: Ω (Na 2 CO 3) \u003d 38.7%

    1) 2NA + 2H 2 O → 2NAOH + H 2
    2) Na 2 O + H 2 O → 2naoh
    1) we find the numberNa.
    n (H 2) \u003d V / VM \u003d 4.48 / 22.4 \u003d 0.2 mol
    By equation (1) n (Na) \u003d N 1 (NaOh) \u003d 2. N (H 2) \u003d 0.4 mol
    2) We will find a massNa.
    M (Na) \u003d n. M \u003d 0.4. 23 \u003d 9.2 g
    3)     we find the numberNaoh.
    n (NaOH) \u003d M (P-P). Ω / M (NaOH) \u003d 240. 0.1 / 40 \u003d 0.6 mol
    4) we find the numberNa. 2 O.
    n (Na 2 O) \u003d (n (NaOH) - N 1 (NaOh)) / 2 \u003d 0.1 mol
    M (Na 2 O) \u003d n. M \u003d 0.1. 62 \u003d 6.2g
    5) findm.assova Dolia.Na.
    Ω (Na) \u003d M (Na) / M (mixtures). 100% \u003d 59.7%
    m (mixtures) \u003d m (na) + m (na 2 O) \u003d 15.4
    Answer: Ω (Na) \u003d59,7%

    H 2 SO 4 + Na 2 CO 3 → Na 2 SO 4 + CO 2 + H 2
    1) Find a total numberH 2 SO 4
    n 1 (H 2 SO 4) \u003d m. ω / m \u003d 490. 0.4 / 98 \u003d 2 mole
    M (H 2 SO 4) \u003d 2 + 32 + 64 \u003d 98 g / mol
    2) FindH 2 SO 4rearachingfrom Na 2 CO 3
    n (Na 2 CO 3) \u003d n (Na 2 CO 3. 10H 2 O) \u003d M / M \u003d 143/286 \u003d 0.5 mol
    M (Na 2 CO 3. 10H 2 O) \u003d (46 + 12 + 48) + (10. 18) \u003d 286 g / mol
    By equation N (H 2 SO 4) \u003d N (Na 2 CO 3) \u003d 0,5mol
    3) findH. 2 SO. 4 Rearamed S.Naoh.
    n (H 2 SO 4) \u003d N 1 -N \u003d 2-0.5 \u003d 1.5mol
    4) Find a lot of NaOH
    n (NaOH) \u003d 2N (H 2 SO 4) \u003d 2. 1,5 \u003d 3 mole
    M (NaOH) \u003d n. M \u003d 3. 40 \u003d 120 g
    M (NaOH) \u003d 23 + 16 + 1 \u003d 18 g / mol
    5) we find a mass shareNaoh.
    Ω (NaOH) \u003d m (ML) /. M (P-P) * 100% \u003d 120/1200. 100% \u003d 10%
    Answer: M (NaOH) \u003d 120g; Ω (NaOH) \u003d 10%

    48) A mixture of magnesium and aluminum sawdust was treated with an excess of diluted hydrochloric acid, while 11.2 liters (N.O.) of hydrogen were separated. If the same mass of the mixture is treated with an excess solution of potassium hydroxide, then 6.72 liters (N.O.) of hydrogen are separated. Calculate the mass fraction of magnesium in the initial mixture.
    49) Calcium carbide weighing 6.4 g was dissolved in 87 ml of bromomic acid (ρ \u003d 1.12 g / ml) with a mass fraction of 20%. What is the mass proportion of bromomomodor in the resulting solution?

 

Perhaps it will be useful to read: