Mole Concept and Stoichiometry — Question 8

$\begin{matrix} 2\text{NH}_4\text{Cl} & + \text{ Ca(OH)}_2 \longrightarrow \text{CaCl}_2 \space + \space 2\text{H}_2\text{O} \space + & 2\text{NH}_3 \\ 2[14 + 4(1) + 35.5] & & 2[14 + 3(1)] \\ 107 \text{g} & &34\text{g} \\ & &2\text{ mole} \\ \end{matrix}$

(a) 107 g NH₄Cl gives 34 g of NH₃

∴ 21.4 g NH₄Cl will give $\dfrac{34}{107}$ x 21.4

= 6.8 g of NH₃

(b) Volume of ammonia produced by 107 g NH₄Cl = 2 x 22.4 L

∴ Volume of ammonia produced by 21.4 g NH₄Cl = $\dfrac{2 \times 22.4}{107}$ x 21.4

= 8.96 L