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ICSE Class 10 Chemistry Question 9 of 28

Stoichiometry — Percentage Composition — Empirical & Molecular Formula — Chemical Equations — Question 17

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17
Question

Question 1(2005)

The equations given below relate to the manufacture of sodium carbonate [Mol. wt. of Na2CO3 = 106]

(i) NaCl + NH3 + CO2 + H2O ⟶ NaHCO3 + NH4Cl

(ii) 2NaHCO3 ⟶ Na2CO3 + H2O + CO2

Questions (a) and (b) are based on the production of 21.2 g. of sodium carbonate.

(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g. of sodium carbonate [Molecular weight of NaHCO3 = 84].

(b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at s.t.p., would be required.

Answer

2NaHCO3⟶Na2CO3+H2O+CO22mol.1mol.2×84106 g.=168 g.\begin{matrix} 2\text{NaHCO}_3 & \longrightarrow & \text{Na}_2\text{CO}_3 & + & \text{H}_2\text{O} + \text{CO}_2 \\ 2 \text{mol.} & & 1 \text{mol.} \\ 2 \times 84 & & 106\text{ g.} \\ = 168\text{ g.} & & \end{matrix}

106 g of Na2CO3 is obtained from 168 g of NaHCO3

∴ 21.2 g. of Na2CO3 is obtained from 168106\dfrac{168}{106} x 21.2 = 33.6 g of NaHCO3

(ii)

NaCl+NH3+CO2+H2O1mol.22.4lit.⟶NaHCO3+NH4Cl1mol.84 g.\begin{matrix} \text{NaCl} & + & \text{NH}_3 & + & \text{CO}_2 & + & \text{H}_2\text{O} \\ & & & & 1\text{mol.} & \\ & & & & 22.4 \text{lit.} \\ \longrightarrow & \text{NaHCO}_3 & + & \text{NH}_4\text{Cl} \\ & 1 \text{mol.} \\ & 84\text{ g.} \end{matrix}

84 g of NaHCO3 is obtained from 22.4 lit of CO2

∴ 33.6 g. of NaHCO3 is obtained from 22.484\dfrac{22.4}{84} x 33.6 = 8.96 lit.

Hence, 8.96 lit of CO2 is required.