Answer:
The hydrogen produces the smaller amount of ammonia.
Step-by-step explanation:
We are given the masses of two reactants, so this is a limiting reactant problem.
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 28.02 2.016 17.03
N₂ + 3H₂ ⟶ 2NH₃
Mass/g: 70.0 7.00
1. Calculate the moles of N₂ and H₂
Moles N₂ = 70.0 × 1/28.02
Moles N₂ = 2.498 mol N₂
Moles H₂ = 7.00 × 2.016
Moles H₂ = 3.472 mol N₂
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2. Calculate the moles of NH₃ from each reactant
From N₂:
The molar ratio is 2 mol NH₃/1 mol N₂
Moles of NH₃ = 2.498 × 2/1
Moles of NH₃ = 4.996 mol NH₃
From H₂:
The molar ratio is 2 mol NH₃/3 mol H₂
Moles of NH₃ = 3.472 × 2/3
Moles of NH₃ = 4.139 mol NH₃
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3. Identify the limiting reactant
The limiting reactant is H₂, because it produces fewer moles of NH₃.
