Calcium carbide, $\text{CaC}_2$, reacts with water, as illustrated. The information below the equation gives, in $\text{kJ mol}^{-1}$, the standard enthalpies of formation for the compounds taking part. $\text{CaC}_2(s) + \text{H}_2\text{O}(l) \rightarrow \text{CaO}(s) + \text{C}_2\text{H}_2(g)$ What is the standard enthalpy change for the reaction shown?
- A$-753\ \text{kJ mol}^{-1}$
- B$-61\ \text{kJ mol}^{-1}$
- C$+61\ \text{kJ mol}^{-1}$
- D$+753\ \text{kJ mol}^{-1}$