Контрольные задания > Part C. C1. Write the reaction equations for the following transformations: Ca→ Ca(OH)₂→CaCO₃ → Ca(HCO₃)₂ → CaCl₂ → Ca(NO₃)₂. Consider transition 1 in terms of redox reactions; transitions 4 and 5 – from the standpoint of electrolytic dissociation. C2. Upon interaction of 24.15g of technical sodium, containing 5% impurities, with water, 8.96L of hydrogen (STP) was obtained. Calculate the yield of the reaction product (in %)
Вопрос:

Part C. C1. Write the reaction equations for the following transformations: Ca→ Ca(OH)₂→CaCO₃ → Ca(HCO₃)₂ → CaCl₂ → Ca(NO₃)₂. Consider transition 1 in terms of redox reactions; transitions 4 and 5 – from the standpoint of electrolytic dissociation. C2. Upon interaction of 24.15g of technical sodium, containing 5% impurities, with water, 8.96L of hydrogen (STP) was obtained. Calculate the yield of the reaction product (in %)

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Ответ:

Part C.

C1. Write the reaction equations for the following transformations:

  1. Ca → Ca(OH)₂: Ca + 2H₂O → Ca(OH)₂ + H₂ (Redox reaction, Ca is oxidized, H is reduced)
  2. Ca(OH)₂ → CaCO₃: Ca(OH)₂ + CO₂ → CaCO₃↓ + H₂O
  3. CaCO₃ → Ca(HCO₃)₂: CaCO₃ + H₂O + CO₂ → Ca(HCO₃)₂
  4. Ca(HCO₃)₂ → CaCl₂: Ca(HCO₃)₂ + 2HCl → CaCl₂ + 2H₂O + 2CO₂ (Electrolytic dissociation of HCl is involved)
  5. CaCl₂ → Ca(NO₃)₂: CaCl₂ + 2AgNO₃ → Ca(NO₃)₂ + 2AgCl↓ (Electrolytic dissociation of AgNO₃ is involved)

C2. Upon interaction of 24.15g of technical sodium, containing 5% impurities, with water, 8.96L of hydrogen (STP) was obtained. Calculate the yield of the reaction product (in %)

1. Calculate the mass of pure sodium:

Mass of impurities = 24.15 g * 5% = 24.15 * 0.05 = 1.2075 g

Mass of pure Na = 24.15 g - 1.2075 g = 22.9425 g

2. Calculate the moles of pure sodium:

Molar mass of Na = 22.99 g/mol

Moles of Na = 22.9425 g / 22.99 g/mol ≈ 0.998 mol

3. Write the balanced chemical equation for the reaction of sodium with water:

2Na + 2H₂O → 2NaOH + H₂

4. Calculate the theoretical yield of hydrogen (in moles):

From the stoichiometry, 2 moles of Na produce 1 mole of H₂.

Theoretical moles of H₂ = (0.998 mol Na) / 2 = 0.499 mol H₂

5. Calculate the theoretical yield of hydrogen (in liters):

Molar volume of gas at STP = 22.4 L/mol

Theoretical volume of H₂ = 0.499 mol * 22.4 L/mol ≈ 11.1776 L

6. Calculate the percentage yield:

Actual yield of H₂ = 8.96 L

Percentage yield = (Actual yield / Theoretical yield) * 100%

Percentage yield = (8.96 L / 11.1776 L) * 100% ≈ 80.16%

Ответ: The yield of the reaction product is approximately 80.16%.

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