Solvequill Blog · chemistry · 3 min read · 19 views

28 g of N₂ and 4 g of H₂ — which one runs out first?

Grams cannot be compared directly; moles can. Converting first turns a limiting-reagent question into a single ratio check.

Published: August 25, 2026

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The Solvequill lesson for this question.

The question

28.0 g of nitrogen gas reacts with 4.0 g of hydrogen gas to form ammonia by N₂ + 3H₂ → 2NH₃. Identify the limiting reagent and calculate the mass of ammonia produced. Use $M (N_{2}) = 28.0 \frac{g}{m o l}, M (H_{2}) = 2.0 \frac{g}{m o l}$ and $M (N H_{3}) = 17.0 \frac{g}{m o l}$ .

Hydrogen is short of what the equation demands, so it decides the yield.

What to notice first

Mass tells you nothing about which reagent runs out, because the equation counts PARTICLES, not grams. Convert both to moles, then compare what you have against what the equation demands — the ratio decides, never the larger mass.

Working it through

Convert both masses to moles:

n (N_{2}) = \frac{28.0}{28.0} = 1.00 mol, n (H_{2}) = \frac{4.0}{2.0} = 2.00 mol

The equation demands three moles of hydrogen for every one of nitrogen. Ask what 1.00 mol of N₂ would need:

n (H_{2})_{needed} = 3 \times 1.00 = 3.00 mol

Only 2.00 mol is available, so hydrogen runs out first — despite there being seven times more nitrogen by mass. Hydrogen is the limiting reagent, and it alone sets the yield:

n (N H_{3}) = 2.00 \times \frac{2}{3} = 1.33 mol

Convert the product back to grams:

m (N H_{3}) = 1.33 \times 17.0

The answer

m (N H_{3}) \approx 22.7 g

Only 0.67 mol of the nitrogen is consumed, leaving about 9.3 g of N₂ unreacted. A limiting-reagent answer is incomplete without noticing that the excess reagent is still sitting there.

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