Final answer:
The mole ratio of hydrogen gas to ammonia gas is 3:2. The limiting reagent is nitrogen gas. The theoretical yield is 166.4 grams of ammonia and the percent yield is 58.5%.
Step-by-step explanation:
The balanced equation for the synthesis of ammonia is N2 + 3H2 → 2NH3.
a) The mole ratio of hydrogen gas to ammonia gas in the balanced reaction is 3:2. This means that for every 3 moles of H2, 2 moles of NH3 are produced.
b) To determine the limiting reagent, compare the moles of reactants to the mole ratio in the balanced equation. In this case, we have 450 grams of N2, which is equivalent to 15.9 moles, and 80 grams of H2, which is equivalent to 40 moles. Since the mole ratio is 3:2, we can see that the limiting reagent is N2.
c) The theoretical yield can be calculated by multiplying the moles of the limiting reagent (15.9 moles of N2) by the mole ratio (2 moles of NH3 / 3 moles of N2), which gives 10.6 moles of NH3. Converting moles to grams (using the molar mass of NH3), the theoretical yield is 166.4 grams of NH3.
d) The percent yield is the actual yield (given in the question) divided by the theoretical yield, multiplied by 100%. In this case, if the actual yield is 6.2 moles of NH3 (which is equivalent to 170 grams), the percent yield would be (6.2 moles / 10.6 moles) x 100% = 58.5%.