Start by balancing the chemical equation for this reaction.
3H2 + N2 --> 2NH3
I am going to show how to set up the calculation one piece at a time. After setting up all the pieces, I will evaluate the expression in one calculation in order to avoid rounding errors.
Start by writing down the given, 120g H2. Convert to moles using the molar mass of H2 (2.02 g/mol). Think of the molar mass as a ratio and arrange it such that the units of g H2 cancel. In this case, use the reciprocal mol/g ratio. Notice that the number 2.02 stays with the unit grams when writing the reciprocal.
120g H2 x [(1 mole H2)/(2.02g H2)]
The g H2 units cancel and the calculation is in the unit of mole H2 at this point. Next, use the stoichiometric coefficients from the balanced chemical reaction for the mole-to-mole ratio of hydrogen to ammonia (3 moles H2)/(2 moles NH3). Arrange the ratio so that the moles H2 in the ratio cancels with the mole H2 already in the calculation. For the current calculation, mole H2 appears in the numerator, so the next step requires mole H2 in the denominator. Therefore, use the ratio (2 moles NH3)/(3 moles H2).
120g H2 x [(1 mole H2)/(2.02g H2)] x [(2 moles NH3)/(3 moles H2)]
The mole H2 units cancel and now the calculation is in unit of moles NH3. We're almost finished. This value for ammonia needs to be expressed in grams, and you can convert moles to grams by multiplying by the molar mass of ammonia (17.03g/mol).
120g H2 x [(1 mole H2)/(2.02g H2)] x [(2 moles NH3)/(3 moles H2)] x [(17.03g NH3)/(1 mole NH3)]
Evaluate this expression and write the answer with 2 significant figures (because 120g was the starting value and has 2 significant figures).
674.5g NH3 rounds to 670g NH3.
