Question

Given a combustion reaction: $C_6{H}_{6\left(l\right)}+\frac{15}{2}{O}_{2\left(g\right)}\to 6C{O}_{2\left(g\right)}+3H_2{O}_{\left(l\right)},\Delta H=-3264$ kJ/mol

The energy (in kJ) evolved when $3.9$ g of benzene is burnt in air will be:

• A. $32.46$
• B. $16.32$
• C. $326.4$
• D. $163.2$

Answer 1

The correct option is D $163.2$

The combustion reaction is as follows:

$C_6{H}_{6\left(l\right)}+\frac{15}{2}{O}_{2\left(g\right)}\to 6C{O}_{2\left(g\right)}+3H_{2}O\left(l\right),\Delta H=-3264$ kJ/mol

$3.9$ g of benzene $=\frac{3.9}{78}$ mol $=\frac{1}{20}$ mol

For $1$ mol of $C_6{H}_6\left(l\right)$, energy evolved $=-3264$ kJ

For $\frac{1}{20}$ mol of $C_6{H}_6\left(l\right)$ , energy evolved $=\frac{1}{20}\times -3264$ $=-163.2$ kJ

Therefore, $163.2$ kJ of heat will be released.