A gaseous mixture of ethene, ethane and methane having total volume 150 ml is subjected to combustion in excess of oxygen. If percentage of methane in the original mixture is 20%, then calculate volume (in ml of ${CO}_{2} (g)$ which will be obtained at same temperature and pressure.

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240 ml

30 ml

270 ml

150 ml

answer is D.

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Detailed Solution

$\underset{30 ml}{{CH}_{4}} \underset{120 ml}{\underset{⏟}{C_{2} H_{4} C_{2} H_{6}}}$

${\underset{30 ml}{{CH}_{4}}}_{} + 2 O_{2} ⟶ \underset{30 ml}{{CO}_{2}} (g) + 2 H_{2} O$

$\underset{x}{C_{2} H_{4}} + O_{2} ⟶ \underset{2 x}{2 {CO}_{2}}$

$\underset{120 - x}{C_{2} H_{6}} + O_{2} ⟶ \underset{240 - 2 x}{2 {CO}_{2}}$

$V_{{CO}_{2}} = 30 + 2 x + 240 - 2 x$

$= 270 ml$

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