Nitroglycerine $(MW = 227.1)$ detonates according to the following equation:
$2 C_{3}H_{5}(NO_{3})_{3(l)} \longrightarrow 3 N_{2(g)} + \frac{1}{2} O_{2(g)} + 6 CO_{2(g)} + 5 H_{2}O_{(g)}$
The standard molar enthalpies of formation,$\Delta H_{f}^{\circ}$ for all the compounds are given below:
$\Delta H_{f}^{\circ}[C_{3}H_{5}(NO_{3})_{3}] = -364 \, kJ/mol$
$\Delta H_{f}^{\circ}[CO_{2(g)}] = -393.5 \, kJ/mol$
$\Delta H_{f}^{\circ}[H_{2}O_{(g)}] = -241.8 \, kJ/mol$
$\Delta H_{f}^{\circ}[N_{2(g)}] = 0 \, kJ/mol$
$\Delta H_{f}^{\circ}[O_{2(g)}] = 0 \, kJ/mol$
The enthalpy change when $10 \, g$ of nitroglycerine is detonated is $..... \, kJ$.