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ICSE Class 10 Chemistry Question 1 of 28

Stoichiometry — Percentage Composition — Empirical & Molecular Formula — Chemical Equations — Question 3

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Question 3

From the equation : 3Cu + 8HNO3 ⟶ 3Cu(NO3)2+ 4H2O + 2NO. Calculate

(i) the mass of copper needed to react with 63 g of nitric acid

(ii) the volume of nitric oxide collected at the same time. [Cu = 64, H = 1, O = 16, N = 14]

Answer

3Cu+8HNO3⟶3Cu[NO3]2+4H2O+2NO3(64)8[1+14+48]=192 g=8×63=504 g\begin{matrix} 3\text{Cu} & + & 8\text{HNO}_3 & \longrightarrow & 3\text{Cu[NO}_3]_2 & + & 4\text{H}_2\text{O} & + & 2\text{NO} \\ 3(64) & & 8[1 + 14 + 48] \\ = 192 \text{ g} & & = 8 \times 63 = 504 \text{ g} \\ \end{matrix}

504 g of nitric acid reacts with 192 g of copper

∴ 63 g of nitric acid will react with 192504\dfrac{192}{504} x 63 = 24 g of copper

Hence, 24 g of copper is needed to react with 63 g of nitric acid

(ii)

3Cu+8HNO3504 g⟶3Cu[NO3]2+4H2O+2NO2(22.4)lit.3\text{Cu} + \underset{504 \text{ g}}{8\text{HNO}_3} \longrightarrow 3\text{Cu[NO}_3]_2 + 4\text{H}_2\text{O} + \underset{2(22.4) \text {lit.}}{2\text{NO}}

504 g of nitric acid liberates (2 x 22.4) lit of Nitric Oxide

∴ 63 g of nitric acid will liberate 2×22.4504\dfrac{2 \times 22.4}{504} x 63 = 5.6 lits of Nitric Oxide

Hence, 5.6 lits. of nitric oxide is collected.