Is the reactant with the smaller mass always the limiting reagent?
Short answer: Compare moles to the ratio: 1 mol N₂ needs 3 mol H₂, and we have 5 mol H₂. So N₂ runs out first, despite its bigger mass.
The trap
The wrong idea
“The reactant with the smaller mass is always the limiting reagent.”
It sounds reasonable, and a lot of JEE/NEET aspirants carry it into the exam. Here's a one-question check, then the explanation that makes it click.
Quick check
Try it before you peek
N₂ + 3H₂ → 2NH₃. With 28 g N₂ and 10 g H₂, the limiting reagent is:
- H₂ (smaller mass)
- N₂
- Neither
Show the answer
B. N₂
Compare moles to the ratio: 1 mol N₂ needs 3 mol H₂, and we have 5 mol H₂. So N₂ runs out first, despite its bigger mass.
The click
Reactions work in moles, in the ratio set by the balanced equation, not in grams. The limiting reagent is the reactant that runs out first, and you can only see which one that is after converting masses to moles.
For N₂ + 3H₂ → 2NH₃: moles of N₂ = 28 g ÷ 28 g/mol = 1 mol, and moles of H₂ = 10 g ÷ 2 g/mol = 5 mol. The equation says 1 mol of N₂ needs 3 mol of H₂. We have 5 mol of H₂, more than the 3 mol needed, so H₂ is in excess and N₂ runs out first. N₂ is the limiting reagent, even though its mass (28 g) is the larger one.
The limiting reagent decides how much product forms. Assuming the reaction goes to completion, 1 mol of N₂ gives 2 mol of NH₃ = 2 × 17 g = 34 g, and 5 − 3 = 2 mol (4 g) of H₂ is left over. Mass check: 28 g + 6 g of reactants used = 34 g of product.
A fast method: divide the moles of each reactant by its coefficient in the balanced equation. The smallest result marks the limiting reagent. Here N₂ gives 1 ÷ 1 = 1 and H₂ gives 5 ÷ 3 ≈ 1.67, so N₂ is limiting.
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