ChemicalStoichiometrymediumFE ExamCoursework

Maximum Ammonia from a Non-Stoichiometric Feed

Ammonia is produced by

$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$

A reactor is fed $100\ \text{mol}$ of $\text{N}_2$ and $240\ \text{mol}$ of $\text{H}_2$ .

Assuming complete conversion of the limiting reactant, find the maximum moles of $\text{NH}_3$ produced.

Given

nN2	100 mol	Nitrogen fed
nH2	240 mol	Hydrogen fed

Hint 1

Do not assume the smaller feed is limiting — the coefficients matter.

Hint 2

Divide each feed by its stoichiometric coefficient; the smallest ratio limits.

Hint 3

N₂: 100/1 = 100. H₂: 240/3 = 80. Which is smaller?

Worked solution — try the problem first

Identify the limiting reactant. Compare what is available against what the stoichiometry demands.

To consume all 100 mol N₂ would require

$100 \times 3 = 300\ \text{mol H}_2$

but only 240 mol is available. Hydrogen is limiting; nitrogen is in excess.

Product from the limiting reactant. From the equation, 3 mol H₂ yields 2 mol NH₃:

$n_{NH_3} = 240 \times \frac{2}{3} = 160\ \text{mol}$

$\boxed{n_{NH_3} = 160\ \text{mol}}$

Check the excess. H₂ consumed: 240 mol, requiring $240/3 = 80$ mol N₂. So 20 mol N₂ leaves unreacted — consistent with nitrogen being in excess.

A faster test. Divide each feed by its stoichiometric coefficient and take the smallest: N₂ gives $100/1 = 100$ , H₂ gives $240/3 = 80$ . Hydrogen's 80 is smaller, so it limits, and the extent of reaction is 80. Then $n_{NH_3} = 2 \times 80 = 160$ mol.

Concepts:Limiting reactantStoichiometryExtent of reaction

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