From the equation

3Cu + 8HNO₃ ⟶ 3Cu(NO₃)₂ + 4H₂O + 2NO

(At. mass Cu=64, H=1, N=14, O=16)

Calculate:

(a) Mass of copper needed to react with 63 g of HNO₃

(b) Volume of nitric oxide at S.T.P. that can be collected.

$\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.