Given the following:  4 Al(s) + 3O2 (g) --> 2 Al2O3 (s)    ΔrxnH= -3202 kJ
    Determine the volume of water present in a calorimeter if 20.0 g of aluminum are burned in the reaction chamber, resulting in a increase of 15.7°C.

The answer given: Vwater= 9.02 L

convert g of Al to moles of Al
now from the reactions 4 moles of Al ---> 3202KJ of energy
get the amount of energy for the calculated moles of Al
This amount is q
now q = m c delta t -----(1)
m = mass of water = kg  of water as density of water is 1Kg/L

now you have c of water = 4.18KJ /Kg .deg C
delta t = 15.7 deg C so plug in value of q , delta t and c  in eq 1 to m of water