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ICSE Class 10 Chemistry Question 2 of 28

Stoichiometry — Percentage Composition — Empirical & Molecular Formula — Chemical Equations — Question 9

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Question 9

Sodium carbonate reacts with dil. H2SO4 to give the respective salt, water and carbon dioxide. Calculate the mass of pure salt formed when 300 g. of Na2CO3 of 80% purity reacts with dil. H2SO4. [Na = 23, C = 12, O = 16, H = 1, S = 32].

Answer

Given,

300 g. of Na2CO3 is of 80% purity

∴ Mass of pure Na2CO3 present in it = 80100\dfrac{80}{100} x 300 = 240 g

We have,

Na2CO3+H2SO4⟶Na2SO4+H2O+CO2(2×23)+12(2×23)+32+(3×16)+(4×16)=106 g=142 g\begin{matrix} \text{Na}_2\text{CO}_3 & + & \text{H}_2\text{SO}_4 & \longrightarrow & \text{Na}_2\text{SO}_4 & + & \text{H}_2\text{O} & + & \text{CO}_2 \\ (2 \times 23) + 12 & & & & (2 \times 23) + 32 \\ + (3 \times 16 ) & & & & + (4 \times 16) \\ = 106 \text{ g} & & & & = 142 \text{ g} \end{matrix}

106 g of Na2CO3 gives 142 g of Na2SO4

∴ 240 g of Na2CO3 gives 142106\dfrac{142}{106} x 240 = 321.51 g

Hence, mass of pure salt formed is 321.51 g