Calculate the volume of $C O_{2}$ evolved by the combustion of 50 ml of a mixture contains $C_{2} H_{4}$ and $60 %$ $C H_{4}$ (by volume)

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Solution

50 ml mixture of $C_{2} H_{4}$ and $C H_{4},$
where $C H_{4} = 50 % o f 60 m l = 30 m l$ and $C_{2} H_{4} = 20 m l .$
applying limiting reagent law
$C_{2} H_{4} + C H_{4} + 5 O_{2} ⟶ 3 C O_{2} + 4 H_{2} O$
$(20 - x) (30 - x) 3 x$
limiting reagent is $C_{2} H_{4}$
$∴ (20 - x) = 0$ and $x = 20$
putting the value of x for $C O_{2}$
the volume of $C O_{2} = 3 x = 3 \times 20 = 60 m l$

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