Question

The ratio of number of moles of $O_2$ required to the number of moles of $H_{2}O$ produced during complete combustion of 1 mole of benzene is :

• A. $2:5$
• B. $2:3$
• C. $5:2$
• D. $3:2$

Answer 1

The correct option is C $5:2$

Combustion of aromatic compounds gives $C{O}_2$ and $H_{2}O$ with sooty flame.

When benzene is heated in air, it burns with a sooty flame producing $C{O}_2$ and $H_{2}O$.

General reaction for the combustion of any hydrocarbon
$C_x{H}_y+(x+\frac{y}{4})O_2\to xC{O}_2+\left(\frac{y}{2}\right)H_{2}O$
For benzene, x = 6 and y = 6
Therefore for 1 mole of benzene,
$C_6{H}_6+\left(\frac{15}{2}\right)O_2\to 6C{O}_2+3H_{2}O$
Thus, 1 mole of benzene will react with $\frac{15}{2}$ mole of $O_2$ to give 6 moles of $C{O}_2$ and 3 mole of $H_{2}O$
Required ratio:
$\frac{{O_2}_{reqd.}}{{H_{2}O}_{produced}}=\frac{15}{2\times 3}=\frac{5}{2}=5:2$