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) + $21}{2}$ O2 (g) $\stackrel{}{\to }$ 2NO (g) + 3H2O (l) What mass of steam is produced when 1.5 g of nitrogen monoxide is formed?

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Solution

## (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 $×$ Volume of O2 Volume of H2 produced = 2500 cm3 2500 = 2 $×$ 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 = $\frac{3}{2}×0.05=0.075$ Mass of water produced = Number of moles of water $×$ Molar mass of water = (0.075$×$18) g = 1.35 g Hence, mass of water (or steam) produced is 1.35 g.

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