Q.When 22.4 litres of H₂(g) is mixed with 11.2 litres of Cl₂(g), each at 273 K at 1 atm, the moles of HCl(g) formed is equal to

The Intuition: A Sandwich Analogy

Imagine you are 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 complete sandwiches can you make?

You might think: "I have enough bread for 5 sandwiches (10 ÷ 2), and enough cheese for 4 sandwiches (4 ÷ 1)." But you cannot make 5 sandwiches — after the 4th sandwich, you run out of cheese. The cheese stops you. The cheese is the limiting reagent.

The bread is in excess — you have 2 slices left over, but they are useless without cheese.

This is the core idea: in a chemical reaction, reactants are not always present in the exact ratio required by the balanced equation. One reactant runs out first, and when it does, the reaction stops — no matter how much of the other reactants remain.

The Precise Definition

The other reactants are called excess reagents — they are present in amounts greater than needed to react with the limiting reagent.

How to Identify the Limiting Reagent (Step-by-Step)

Consider the reaction:

Suppose you start with 4 moles of $H_2$ and 3 moles of $O_2$.

Step 1: Write the balanced equation and note the mole ratio.

From the equation: 2 moles $H_2$ react with 1 mole $O_2$.

So the required ratio is $\frac{H_2}{O_2} = \frac{2}{1}$.

Step 2: Calculate how much of one reactant is needed to completely use up the other.

• If $O_2$ is the limiting reagent: 3 moles $O_2$ would need $3 \times 2 = 6$ moles $H_2$. But you only have 4 moles $H_2$ — not enough. So $O_2$ cannot be limiting.
• If $H_2$ is the limiting reagent: 4 moles $H_2$ would need $4 \div 2 = 2$ moles $O_2$. You have 3 moles $O_2$ — more than enough. So $H_2$ runs out first.

Conclusion: $H_2$ is the limiting reagent. $O_2$ is in excess (1 mole remains unreacted).

Why It Matters

The limiting reagent directly tells you the theoretical yield — the maximum product possible. In the example above, since $H_2$ is limiting, the amount of $H_2O$ formed is based on $H_2$:

$4 \text{ mol } H_2 \times \frac{2 \text{ mol } H_2O}{2 \text{ mol } H_2} = 4 \text{ mol } H_2O$ …