Suppose you burn 0.300 g of $C$ (graphite) in an excess of ${\mathrm{O}}_2(\mathrm{g})$ in a constant volume calorimeter to give ${\mathrm{CO}}_2(\mathrm{g})$. $\mathrm{C}(\text{ graphite })+{\mathrm{O}}_2(\mathrm{g})⟶{\mathrm{CO}}_2(\mathrm{g})$ The temperature of the calorimeter, which contains $775\mathrm{g}$ of water, increases from $25.00^{\circ }\mathrm{C}$ to $27.38^{\circ }\mathrm{C}.$ The heat capacity of the bomb is $893\mathrm{J}/\mathrm{K}$. What quantity of heat is evolved per mole of carbon?

Step by Step Solution: Step 1. Calculate the number of moles of $C$ used: Step 2. Use the balanced chemical equation to determine the number of moles of ${\mathrm{CO}}_2$ produced. Step 3. Determine the amount of heat absorbed by the water and the calorimeter using the equation $q=mc\Delta T.$ Step 4. Subtract the heat absorbed by the water and the calorimeter from the total heat absorbed by the system to obtain the heat evolved by the combustion of $C$. Step 5. Calculate the heat evolved per mole of $C$ by dividing the heat evolved by the number of moles of $C$ used. Final Answer: The quantity of heat evolved per mole of carbon is -394.4 kJ/mol.