1.20 g of hydrogen gas is completely burned in the presence of excess oxygen gas in a bomb calorimeter producing water. The heat capacity of the calorimeter is 9.43 kJ/°C and the temperature of the calorimeter rose from 25.55 °C to 41.97°C. Calculate the enthalpy change in J and in kJ/mol or water produced.

Question

m=120g
c= 9.43
T1= 25.55 C
T2= 41.97 C

I don't know what to do after this

Answer 1

2H2 + O2 ==> 2H2O
mols H2 gas initially = 1.20/4 = 0.3 and since the ratio of H2O produced to mols H2 initially is 2/2 we must have produced 0.3 mols H2O.

q = heat capacity x (Tfinal-Tinitial) = ? kJ = dH rxn.
So dH rxn in kJ is for 0.3 mol H2O produced so dH rxn/0.3 gives you dH rxn in kJ/1 mol. Then convert to J also.
Post your work if you get stuck.

Answer 2

n=m/M = 1.2/4 =0.3 mols

q=ct = (9.43)(16.42)

I am stuck here