Answer
$\Delta H^{\circ}_{r}$ for the formation of 66.8 g $CO_{2(g)}$ = $2.70 \times 10^{2} kJ/mol$
Work Step by Step
The equation for the decomposition of $CaCO_{3(s)}$ at $850^{\circ}C$ is,
$ CaCO_{3(s)} → CaO_{(s)} + CO_{2(g)}$
$ \Delta H^{\circ}_{f} CaCO_{3(s)} = -1206.9 kJ/mol$
$ \Delta H^{\circ}_{f} CaO_{(s)} = - 635.1 kJ/mol$
$ \Delta H^{\circ}_{f} CO_{2(g)} = -393.5 kJ/mol$
The standard enthalpy of reaction = (sum of standard enthalpies of formation of products) – (sum of standard enthalpies of formation of reactants).
$\Delta H^{\circ}_{r} = ∑H^{\circ}_f{products} - ∑ H^{\circ}_f{reactants}$.
$\Delta H^{\circ}_{r} = {[- 635.1 + - 393.5] – [- 1206.9]} kJ/mol = - 178.3 kJ/mol$
Thus one mole (44 grams) of $CO_{2(g)}$ is formed during the above reaction.
Therefore $\Delta H^{\circ}_{r}$ for the formation of 66.8 g $CO_{2(g)}$ = $(457.2\div44)\times66.8 = 270.69kJ/mol =2.70 \times 10^{2} kJ/mol$
$\Delta H^{\circ}_{r}$ for the formation of 66.8 g $CO_{2(g)}$ = $2.70 \times 10^{2} kJ/mol$
