What volume of air $in m^{3}$ is needed for the combustion of $1 m^{3}$ of a gas having the following composition in percentage volume: $2 %$ of $C_{2} H_{2}$ , $8 %$ of $C O$ , $35 %$ of $C H_{4}$ , $50 %$ of $H_{2}$ and $5 %$ of non-combustible gas. The air contains $20.8 %$ (by volume) of oxygen.

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Solution

The respective oxidation reactions are,
$2 H_{2} + O_{2} \to 2 H_{2} O$
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
$2 C O + O_{2} \to 2 C O_{2}$
$C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 C O_{2} + H_{2} O$
From the given percentage of gases,
Here volume of $C_{2} H_{2} = 0.02 m^{3}$
volume of $O_{2}$ required = $0.05 m^{3}$

volume of $C O = 0.08 m^{3}$
volume of $O_{2}$ required = $0.04 m^{3}$

volume of $C H_{4} = 0.35 m^{3}$
volume of $O_{2}$ required = $0.7 m^{3}$

Volume of $H_{2} = 0.50 m^{3}$
volume of $O_{2}$ required = $0.25 m^{3}$

Hence the total required volume of oxygen for the combustion of given gas mixture $= 1.04 m^{3}$
Hence, volume of air required $= \frac{1.04 \times 100}{20.8} = 5 m^{3}$

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What volume of air in m3 is needed for the combustion of 1 m3 of a gas having the following composition in percentage volume: 2% of C2H2, 8% of CO, 35% of CH4, 50% of H2 and 5% of non-combustible gas. The air contains 20.8% (by volume) of oxygen.