Question

Ammonia is produced by the following reaction. 3H2(g) + N2(g)

2NH3(g) When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because

Answer 1

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.