How do I solve this? What is the answer?

Question

Given the balanced reaction: N2 (g) + 3H2 (g) ⟶ 2NH3 (g) If you had 84 grams of nitrogen gas mixed with 29 grams of hydrogen, what would the maximum amount (in grams) of ammonia that would be theoretically produced?

Answer 1

Nitrogen (N₂) has a molar mass of 14.007 x 2 = 28.014 g/mol so the number of moles of nitrogen is (84 grams)/(28.014 g/mol) = 2.9985 moles

Hydrogen (H₂) has a molar mass of 1.008 x 2 = 2.016 g/mol so the number of moles of hydrogen is (29 grams)/(2.016 g/mol) = 14.385 moles

According to the balanced chemical reaction equation, it takes 3 moles of hydrogen for every 1 mole of nitrogen to produce the ammonia. But there is almost 5 times the amount of hydrogen so we have excess hydrogen. That means the amount of nitrogen will be limiting the reaction.

So, again, in looking at the balanced chemical reaction equation, we see that there are 2 moles of ammonia produced for every 1 mole of nitrogen (the limiting reactant), so since there are 2.9985 moles of nitrogen, 2.9985 x 2 or 5.997 moles of ammonia will be produced.

The molar mass of ammonia is 14.007 (the nitrogen) + 3 times 1.008 (the hydrogen) or 17.031 g/mol and since there are 5.997 moles of ammonia produced, there will be 17.031 x 5.997 or 102,13 grams of ammonia produced.

The "givens" had 2 sig figs, so the answer should have 2 sig figs as well. So 1.0 x 10² grams of ammonia are theoretically produced