Question

Calcium carbide (CaC2) reacts with water to form acetylene (C2H2) and Ca(OH)2.

Answer 1

The answer is -60.57 = -60.6 KJ.
CaC2(s) + 2 H2O(l) ---> Ca(OH)2(s) +C2H2(g) H= -127.2 KJ
Hf C2H2 = 226.77
Hf Ca(OH)2 = -986.2
Hf H2O = -285.83
Now,

add them up. 226.77 - 986.2 + (2*285.83) = -187.77 
Add back the total enthalpy that is given in the question
 -187.77+127.2 = -60.57 

Answer 2

Answer: The enthalpy of the formation of $CaC_2(s)$ is coming out to be -59.89 kJ/mol.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

For the given chemical reaction:

$CaC_2(s)+2H_2O(l)\rightarrow Ca(OH)_2(s)+C_2H_2(g);\Delta H^o_{rxn}=-127.2kJ$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(1\times \Delta H^o_f_{(Ca(OH)_2(s))})]-[(2\times \Delta H^o_f_{(H_2O(l))})+(1\times \Delta H^o_f_{(CaC_2(s))})]$

We are given:

$\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(Ca(OH)_2(s))}=-986.09kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=227.4kJ/mol\\\Delta H^o_{rxn}=-127.2kJ$

Putting values in above equation, we get:

$-127.2=[(1\times (227.4)})+(1\times (-986.09))]-[(2\times (-285.8))+(1\times \Delta H^o_f_{(CaC_2(s))})]\\\\\Delta H^o_f_{(CaC_2(s))}=-59.89kJ/mol$

Hence, the enthalpy of the formation of $CaC_2(s)$ is coming out to be -59.89 kJ/mol.