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Hess' Law

A 3 L mixture...

Question

A $3 L$ mixture of $C H_{4}$ and $C_{2} H_{6}$ at STP requires $9 L$ of $O_{2}$ for complete combustion. The mole ratio of $C H_{4}$ and $C_{2} H_{6}$ in the mixture is:

A

1 : 1

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B

1 : 2

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C

2 : 1

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D

1 : 3

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Solution

The correct option is B $1 : 2$
The balanced chemical reaction for both the reactions are:
$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O$

Applying Gay lussac's Law,
$1 L$ of $C H_{4}$ requires 2 L of $O_{2}$ at STP.
$1 L$ of $C_{2} H_{6}$ requires $\frac{7}{2}$ L of $O_{2}$ at STP.
Let a and b litres of $C H_{4}$ and $C_{2} H_{6}$ present respectively.
So,
$a$ L of $C H_{4}$ requires $2 a$ L of $O_{2}$ at STP.
$b$ L of $C_{2} H_{6}$ requires $\frac{7}{2}$ $b$ L of $O_{2}$ at STP.
Therefore,
$a + b = 3 2 a + \frac{7 b}{2} = 9$
On Solving,
$a = 1 L$
$b = 2 L$
Mole ratio of $C H_{4} : C_{2} H_{6}$ $= 1 : 2$

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