Question

5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :

• A. Butane
• B. Isobutane
• C. Propane
• D. Ethane

Answer 1

The correct option is C Propane

Solution:- (C) Propane

Combustion of alkane-

$C_n{H}_{2n+2}+\left(\frac{3n+1}{2}\right)⟶nC{O}_2+(n+1)H_{2}O$

From the above reaction,

No. of moles of alkane $=1$

No. of moles of oxygen $=\frac{3n+1}{2}$

From ideal gas equation,

$PV=nRT$

At same temperature and pressure,

$V\propto n$

$\Rightarrow V_{\left(\text{alkane}\right)}{n}_{\left(\text{oxygen}\right)}=V_{\left(\text{oxygen}\right)}{n}_{\left(\text{alkane}\right)}.....\left(1\right)$

Given:-

Volume of alkane $=5L$

Volume of oxygen $=25L$

From ${eq}^n\left(1\right)$, we have

$5\times \left(\frac{3n+1}{2}\right)=25\times 1$

$\frac{3n+1}{2}=\frac{25}{5}$

$\Rightarrow 3n+1=5\times 2$

$\Rightarrow 3n=10-1$

$\Rightarrow n=3$

Thus the alkane is $C_3{H}_{2\times 3+2}=C_3{H}_8$, i.e., propane.