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]

$\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 $\dfrac{192}{504}$ x 63 = 24 g of copper

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

(ii)

$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 $\dfrac{2 \times 22.4}{504}$ x 63 = 5.6 lits of Nitric Oxide

Hence, 5.6 lits. of nitric oxide is collected.