I need help with enthalpy change!!

Question

If 2.98 g of N2(g) is converted to NH3(g) according to the following equation, N2(g) + 3H2(g) ----> 2NH3(g) and the change in enthalpy for the reaction is -92.6 kJ/mol, starting from one mole of N2(g), what is the actual enthalpy change starting with 2.98 g N2(g)?

Answer 1

Since the GIVEN enthalpy change for 1 mole of N2 is -92.6 kJ, all we have to do is find out how many moles of N2 is in 2.98 g of N2, and then multiply that by 92.6. Right?

2.98 g N2 x 1 mol N2/28.0 g = 0.106 moles N2

0.106 moles N2 x -92.6 kJ/mole = -9.86 kJ (answer)