Calculate δh∘f for cac2(s):
cac2(s)+2h2o(l)→ca(oh)2(s)+c2h2(g)< br /> δh∘=−127.2kj

Below are I think the data for this problem:

Given the following data: 

Ca (s) + 2 C (graphite) → CaC2 (s) ∆H = -62.8 kJ 
Ca (s) + ½ O¬2 (g) → CaO (s) ∆H = -635.5 kJ 
CaO (s) + H2O (l) → Ca(OH)2 (aq) ∆H = -653.1 kJ 
C2H2 (g) + 5/2 O¬2 (g) → 2 CO2 (g) + H2O (l)∆H = -1300 kJ 
C (graphite) + O¬2 (g) → CO2 (g) ∆H = -393.51 kJ 

Below is the

CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq) + C2H2 (g) 
So what you do is: 
Times the first equation by -1 Second by 1 Third By 1 Fourth by -1 and Fifth by 2 
So This gives us: 
1.CaC2--> Ca+2C 
2.Ca+1/2O2-->CaO 
3.CaO+H2O-->Ca(OH)2 
4.2CO2+H2O-->C2H2+5/2O2 
5.2C+202-->2CO2 
Now you cancel out like terms on either sides of the equation and you end up with 
CaC2 (s) + 2 H2O (l) → Ca(OH)2 (aq) + C2H2 (g) Just what you wanted 
So to calculate ∆H: 
62.8-635.5-653.1+1300-787.02= -712.82

Anions are negative charged ions