Standard enthalpy of formation PLEASE HELP!!!

Question

The ΔHrxn for this reaction is -902 kJ/mol: 4 NH3(g) - 5 O2(g) → 4 NO(g) + 6 H2O Using this information together with the standard enthalpies of formation given below, calculate the standard enthalpy of formation of NO(g). NH3(g) = -45.9 kJ/mol O2 = 0 kJ/mol H2O = -241.8 kJ/mol

Answer 1

ΔHrxn = sum of products * mole - sum of reactants * mole

The moles come from the coefficients in the reaction

-902 = (-241.8 x 6) + (4 x NO) - (-45.9 x 4) + ( 0 x 5)

-902 = (-1450.8 kJ) + (4 x NO) - (-183.6 )

-902 = (-1450.8 kJ) + (4 x NO) + (183.6) (add 1450.8 to both sides)

548.8 = (4 x NO) + (183.6) (subtract 183.6 to both sides)

365.2 = 4 x NO (divide by 4 on both sides)

91.3 kj/mol = NO