Question

A gas mixture of $3$ liters of propane and butane on complete combustion at ${25}^{o}C$ produced $10$ liters of $C{O}_2$ initial composition of the propane & butane in the gas mixture is:

• A. $66.67\%,33.33\%$
• B. $33.33\%,66.67\%$
• C. $50\%,50\%$
• D. $60\%,40\%$

Answer 1

Answer: (A)

$66.67\%,33.33\%$

$C_3{H}_8+5O_2⟶3C{O}_2+4H_{2}O$

So, $2$ liters of propane $\left(C_3{H}_8\right)$ gives $3\left(2L\right)=6$ Liters of $C{O}_2$.

Now,

$C_4{H}_{10}+\frac{13}{2}{O}_2⟶4C{O}_2+5H_{2}O$

$\therefore 1$ liter of butane would give $4\left(1L\right)=4$ Liters of $C{O}_2$.

Combining $2$ reactions,

$2C_3{H}_8+C_4{H}_{10}+\frac{23}{2}{O}_2⟶10C{O}_2+13H_{2}O$

$\therefore$ Volume of propane $=2$ Litres

$\therefore$ Volume of butane $=1$ Litres

Percentage of propane $=\frac{2}{2+1}\times 100=66.67\%$

Percentage of butane $=\frac{1}{2+1}\times 100=33.33\%$