Question

Given the thermochemical equations X2+3Y2⟶2XY3 ΔH1= −380 kJ X2+2Z2⟶2XZ2 ΔH2= −120 kJ 2Y2+Z2⟶2Y2Z ΔH3=−250 kJ Calculate the change in enthalpy for the reaction. (in kJ) 4XY3+7Z2⟶6Y2Z+4XZ2