Q.In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?

Limiting Reactant Stoichiometry

Imagine you're making sandwiches. Each sandwich needs exactly 2 slices of bread and 1 slice of cheese. You have 10 slices of bread and 4 slices of cheese. How many sandwiches can you make?

You can only make 4 sandwiches — because after that, you run out of cheese. The bread doesn't matter anymore; there's still bread left, but no cheese to complete the sandwich. The cheese limits how many sandwiches you can make.

That's the core idea of a limiting reactant.

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The Intuition

In any chemical reaction, reactants are consumed in a fixed ratio (the stoichiometric coefficients). You never have exactly the right amount of each reactant. One reactant will run out first — that's the limiting reactant. The other reactants are in excess — some of them will be left over when the reaction stops.

The limiting reactant determines:

• How much product you can actually make (the theoretical yield)
• When the reaction stops

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The Precise Statement

Limiting reactant: The reactant that is completely consumed in a chemical reaction, limiting the amount of product formed.

Excess reactant(s): The reactant(s) that remain partially unreacted after the limiting reactant is used up.

To identify the limiting reactant, you compare the actual mole ratio of reactants to the required mole ratio from the balanced equation.

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Step-by-Step Method (Exam-Ready)

1. Write and balance the chemical equation.
2. Convert all given quantities to moles. (If given mass, use molar mass; if given volume and concentration, use $n = C \times V$.)
3. For each reactant, calculate how much product it would produce if it were the limiting reactant. The reactant that gives the smallest amount of product is the limiting reactant.
4. Use the limiting reactant to calculate the actual amount of product formed and the amount of excess reactant consumed.

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Worked Example

Problem: $2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$

You have 5.4 g of Al and 21.3 g of Cl$_2$. Which is limiting?

Step 1: Convert to moles.

Moles of Al = $\frac{5.4}{27} = 0.20$ mol

Moles of Cl$_2$ = $\frac{21.3}{71} = 0.30$ mol

Step 2: Use the shortcut.

For Al: $\frac{0.20}{2} = 0.10$

For Cl$_2$: $\frac{0.30}{3} = 0.10$

They are equal — so neither is limiting? Wait, that's a special case. When the ratios are exactly equal, both reactants are completely consumed. No excess. But here, check carefully:

Step 3: Calculate product from each.

From Al: $0.20 \text{ mol Al} \times \frac{2 \text{ mol AlCl}_3}{2 \text{ mol Al}} = 0.20$ mol AlCl$_3$

From Cl$_2$: $0.30 \text{ mol Cl}_2 \times \frac{2 \text{ mol AlCl}_3}{3 \text{ mol Cl}_2} = 0.20$ mol AlCl$_3$

Both give the same product — so neither is limiting. This is a stoichiometric mixture. Both reactants are used up completely.

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What If They Weren't Equal?

Suppose you had 0.20 mol Al and 0.40 mol Cl$_2$.

Shortcut: Al = 0.10, Cl$_2$ = 0.133. Al is smaller → Al is limiting.

Product from Al: 0.20 mol AlCl$_3$.

Cl$_2$ consumed: $0.20 \text{ mol Al} \times \frac{3 \text{ mol Cl}_2}{2 \text{ mol Al}} = 0.30$ mol Cl$_2$ …