Question

Water can be split into hydrogen and oxygen under suitable conditions. The equation representing the change is:
2H₂O (l) $\stackrel{}{\to }$ 2H₂ (g) + O₂ (g)
(i) If a given experiment results in 2500 cm³ of hydrogen being produced, what volume of oxygen is liberated at the same time under the same conditions of temperature and pressure?
(ii) Ammonia burns in oxygen and the combustion in the presence of a catalyst may be represented as
2NH₃ (g) + $2\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ O₂ (g) $\stackrel{}{\to }$ 2NO (g) + 3H₂O (l)
What mass of steam is produced when 1.5 g of nitrogen monoxide is formed?

Answer 1

(i) 2H₂O (l) $\to$ 2H₂ (g) + O₂ (g)
According to the above balanced equation,
The number of moles of H₂ produced is double the number of moles of O₂ produced.
Hence, the volume of H₂ produced will also be double the volume of O₂ produced.
Volume of H₂ = 2 $\times$ Volume of O₂
Volume of H₂ produced = 2500 cm³
2500 = 2 $\times$ Volume of O₂
Volume of O₂ = 1250 cm³

(ii) Molar mass of nitrogen monoxide (NO) = (14 + 16) g = 30 g
Number of moles in 1.5 g of nitrogen monoxide = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{NO}}{\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{NO}}=\frac{1.5}{30}=0.05\mathrm{mol}$
According to the balanced chemical equation,
For every two moles of nitrogen monoxide produced, three moles of water is produced.
Moles of water produced with 0.05 moles of nitrogen monoxide = $\frac{3}{2}\times 0.05=0.075$
Mass of water produced = Number of moles of water $\times$ Molar mass of water = (0.075$\times$18) g = 1.35 g
Hence, mass of water (or steam) produced is 1.35 g.