Question

N2(g) + 3 H2(g) ---> 2 HN3(g) (blanced)

If 5.42 g of nitrogen are reacted with 5.42 g of hydrogen gas, which of the reactants is the limiting reactant?

Molar Mass of N2 = 28.02 g/mol
Molar Mass of H2 = 2.02 g/mol
Molar Mass of NH3 = 17.04 g/mol

Answer 1

N₂

Step-by-step explanation:

The limiting reactant is the one that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 28.02 2.02 17.04

N₂ + 3H₂ ⟶ 2NH₃

m/g: 5.42 5.42

1. Calculate the moles of each reactant

`\text{Moles of N}_(2) = \text{5.42 g} * \frac{\text{1 mol}}{\text{28.02 g}} = \text{0.1934 mol N}_(2)\\\\\text{Moles of H}_(2) = \text{5.42 g} * \frac{\text{1 mol}}{\text{2.02 g}} = \text{2.683 mol H}_(2)`

2. Identify the limiting reactant

Calculate the moles of NH₃ we can obtain from each reactant.

From N₂:

The molar ratio of NH₃:N₂ is 2:1.

`\text{Moles of NH}_(3) = \text{0.1934 mol N}_(2) * \frac{\text{2 mol NH}_(3)}{\text{1 mol N}_(2)} = \text{0.3867 mol NH}_(3)`

From H₂:

The molar ratio of NH₃:H₂ is 3:2.

`\text{Moles of NH}_(3) = \text{2.683 mol H}_(2) * \frac{\text{2 mol NH}_(3)}{\text{3 mol N}_(2)} = \text{1.789 mol NH}_(3)`

N₂ is the limiting reactant, because it gives the smaller amount of NH₃.