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Molarity

A mixture of ...

Question

A mixture of methane and ethylene in the volume ratio $a : b$ has a total volume of $30$ mL. On complete combustion, it gave $40$ mL of $C O_{2}$ . If the ratio had been $b : a$ instead of $a : b$ , the volume (in $m L$ ) of $C O_{2}$ obtained would be :

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Solution

Volume of mixture $= 30$ mL, $C H_{4} : C_{2} H_{4} :: a : b$
$∴$ Volume of $C H_{4} = \frac{30 a}{(a + b)}$ and $C_{2} H_{4} = \frac{30 a}{(a + b)}$
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
$\frac{30 a}{a + b} 0$ Initial
$0 \frac{30 a}{(a + b)}$ Final
$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$
$\frac{30 b}{a + b} 0$ Initial
$0 \frac{2 \times 30 b}{(a + b)}$ Final
$∴ \frac{30 a}{(a + b)} + \frac{60 b}{a + b} = 40$
$∴ 10 a = 20 b$
$∴ \frac{a}{b} = \frac{2}{1}$
$∴ C H_{4} = 20 m L, C_{2} H_{4} = 10 m L$
If ratio is reversed i.e., $C H_{4} = 10 m L, C_{2} H_{4} = 20 m L$
Then, $C O_{2}$ formed $=$ $10 + 40$ $=$ $50 m L$

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