Answer:
Nâ‚‚
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: Â 28.02 Â 2.02 Â Â 17.04 Â
       N₂ + 3H₂ ⟶ 2NH₃
m/g: Â Â 5.42 Â 5.42
1. Calculate the moles of each reactant Â
[tex]\text{Moles of N}_{2} = \text{5.42 g} \times \dfrac{\text{1 mol}}{\text{28.02 g}} = \text{0.1934 mol N}_{2}\\\\\text{Moles of H}_{2} = \text{5.42 g} \times \dfrac{\text{1 mol}}{\text{2.02 g}} = \text{2.683 mol H}_{2}[/tex]
2. Identify the limiting reactant Â
Calculate the moles of NH₃ we can obtain from each reactant.
From Nâ‚‚: Â
The molar ratio of NH₃:N₂ is 2:1.
[tex]\text{Moles of NH}_{3} = \text{0.1934 mol N}_{2} \times \dfrac{\text{2 mol NH}_{3}}{\text{1 mol N}_{2}} = \text{0.3867 mol NH}_{3}[/tex]
From Hâ‚‚:
The molar ratio of NH₃:Hâ‚‚ is 3:2. Â
[tex]\text{Moles of NH}_{3} = \text{2.683 mol H}_{2} \times \dfrac{\text{2 mol NH}_{3}}{\text{3 mol N}_{2}} = \text{1.789 mol NH}_{3}[/tex]
N₂ is the limiting reactant, because it gives the smaller amount of NH₃.
