nitrogen (N2) gas and hydrogen (H2) gas react to form ammonia (NH3) gas. suppose you have 11.0 mol of N2 and 7.0 mol of H2 in a reactor
Calculate the largest amount of NH3 that could be produced. Round answer to nearest 0.1 mol
Chemistry ?
2020-09-28 2:22 pm
回答 (1)
2020-09-28 2:30 pm
N₂(g) + 3H₂(g) ⇌ 2NH₃(aq)
Mole ratio N₂ : H₂ = 1 : 3
If H₂ completely reacts, N₂ needed = (7.0 mol) × (1/3) = 2.33 mol < 11.0 mol
N₂ is in excess, and H₂ is the limiting reactant.
According to the above equation, mole ratio H₂ : NH₃ = 3 : 2
Moles of NH₃ produced = (7 mol) × (2/3) = 4.7 mol
Mole ratio N₂ : H₂ = 1 : 3
If H₂ completely reacts, N₂ needed = (7.0 mol) × (1/3) = 2.33 mol < 11.0 mol
N₂ is in excess, and H₂ is the limiting reactant.
According to the above equation, mole ratio H₂ : NH₃ = 3 : 2
Moles of NH₃ produced = (7 mol) × (2/3) = 4.7 mol
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