Calcium carbide (cac2) reacts with water to form acetylene (c2h2) and ca(oh)2.
from the following enthalpy of reaction data and data in appendix c, calculate δh∘f for cac2(s):
cac2(s)+2h2o(l)→ca(oh)2(s)+c2h2(g)δ h∘=−127.2kj
δh∘f = kj?

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 

Coal i believe hope this