The $\mathrm{\Delta }{H}_f^{\circ }$of MgO is $-602{\mathrm{kJ}}^2\times {\mathrm{mol}}^{-1}$When 20.15 g MgO is decomposed at constant pressure according to the equation below, how much heat will be transferred?

$2\mathrm{MgO}\left(s\right)⟶2\mathrm{Mg}\left(s\right)+{\mathrm{O}}_2\left(g\right)$

Number of moles = $\begin{array}{r}=\frac{\text{ Mass }}{\text{ Molar mass }}\end{array}$

                                    $\begin{array}{l}=\frac{20.15\mathrm{g}}{40.3\mathrm{g}/\mathrm{mol}}\\ =0.5\mathrm{mol}\end{array}$

Therefore, when $0.5\mathrm{mol}$ of $\mathrm{MgO}$ decomposes, the amount of heat is calculated as follows:

Amount of heat =$\begin{array}{l}(\mathrm{\Delta }H)=\frac{1204\mathrm{kJ}}{2\mathrm{molMgO}\left(\mathrm{s}\right)}\times 0.5\mathrm{molMgO}\left(\mathrm{s}\right)\end{array}$

                            $=301 \mathrm{kJ}$.