Mole Concept & Stoichiometry Miscellaneous Exercises — Question 2

$\begin{matrix} 3\text{Cu} & + &8\text{HNO}_3 & \longrightarrow & 3\text{Cu(NO}_3)_2 + 4\text{H}_2\text{O} + 2\text{NO} \\ 3(64) && 8(1 + 14 + 3(16)) \\ = 192 \text{g} & & = 8(63) \\ &&= 504 \text{g} \\ \end{matrix}$

(a) 504 g nitric acid reacts with 192 g of copper

∴ 63 g of nitric acid reacts with $\dfrac{192}{504}$ x 63 = 24 g of copper

Hence, 24 g of copper is required.

(b) 504 g of nitric acid gives 2 × 22.4 litre volume of NO

∴ 63 g of nitric acid gives $\dfrac{2 × 22.4}{504}$ x 63

= 5.6 litre of NO

5.6 L of NO is collected.