Answer:
49671 L is the produced volume of ammonia
Explanation:
We think the reaction of ammonia 's production:
N₂(g) + 3H₂(g) → 2NH₃ (g)
We have the moles of each reactant so let's determine the limiting reactant:
Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂
Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂
We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂
1 mol of N₂ can produce 2 moles of ammonia
Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃
We apply now, the Ideal Gases Law → P . V = n . R .T
V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm
V = 49671 L
We confirm that the nitrogen was the limiting reactant
3 moles of H₂ need 1 mol of nitrogen to react
Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂
It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles
