Answer
$\Delta H = -196.1kJ$
Work Step by Step
$\Delta H^o_f : NH_3(g) = -46.19 kJ/mol$
$\Delta H^o_f : O_2(g) = 0 kJ/mol$
$\Delta H^o_f : N_2H_4(g) = 95.4 kJ/mol$
$\Delta H^o_f : H_2O (l)= -285.83 kJ/mol$
$4 NH_3(g) + O_2(g) --> 2N_2H_4(g)+ 2H_2O(l)$
$\Delta H^o =(2*(95.4) + 2*(-285.83)) - (4*(-46.19) + 0)$
$\Delta H^o = (190.8 - 571.66) - (-184.76)$
$\Delta H^o = -380.86 + 184.76$
$\Delta H^o = -196.1kJ$
