Hello, could someone please show me how to answer this question:

2N2H4(l)+N2O4(l)=3N2(g)+4H2O(g)

Calculate the enthalpy change for the above reaction using the following data:

H2O(l) = -285kJ mol-1

H2O(g)= -242 kJ mol-1

N2H4(l) = +51 kJ mol-1

N2H4(g) = +95 kJ mol-1

N2O4(l) = -20 kJ mol-1

N2O4(g) = +10 kJ mol-1