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A mixture of methane and ethylene in the volume ratio x:y has a total volume of $30$ ml. On complete combustion it gave $40$ ml of $C O_{2}$ . If the ratio had been y:x, instead of x:y, what volume of $C O_{2}$ could have been obtained?

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$75$ mL

$100$ mL

$25$ mL

$50$ mL

answer is A.

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Detailed Solution

$x (C H_{4})$ + $y (C_{2} H_{4})$ -------> $(x + 2 y) C O_{2}$

x + y = $30$ ml, x + y = $40$ ml

$30$ + y = $40$ ml

y = $10$ ml, x = $20$ ml

Reactants= x+y = $20$ + $10$ = $30$ ml

Products= y+2x = $10$ + $2 (20)$ = $50$ ml

Hence, the correct option is A.

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