Question

Using the information provided calculate the heat of reaction involved in the conversion of 25.00 g of methane (CH4, MM=16.0426 g/mol) to chloroform (CHCl3, MM=119.3779 g/mol):

CH4(g) + 3 Cl2(g) --> CHCl3(l) + 3 HCl(g)

Given the following:
C(graphite) + 2 H2(g) --> CH4(g)
deltaH = -92.3 kJ

C(graphite) + 1/2 H2(g) + 3/2 Cl2(g) --> CHCl3(l)
dH = -74.5 kJ

1/2 H2(g) + 1/2 Cl2(g) --> HCl(g)
dH = -135.1 kJ

ANS:
-603.9 kJ

Much help is appreciated! Thank you!