Mole Concept & Stoichiometry Miscellaneous Exercises — Question 45

$\begin{matrix} 3\text{CuO} & + & 2\text{NH}_3 \\ 3\text{ mol.} & & 2 \text{ mol.} \\ = 3 \times 80 \text{ g.} & & = 2 \times 22.4 \text{ lit.} \\ = 240 \text{ g.} & & = 44.8\text{ lit.} \\ \end{matrix}$

$\longrightarrow 3\text{Cu} + 3\text{H}_2\text{O} + \text{N}_2$

240 g of CuO is reduced by 44.8 lit of NH₃

∴ 120 g of CuO is reduced by $\dfrac{44.8}{240}$ x 120 = 22.4 lit.

Hence, 22.4 lit of NH₃ is required