Limiting reagent problems

WHAT is a limiting reagent?

WHY does it happen? (first principles)

A balanced equation like

$N_2 + 3H_2 \rightarrow 2NH_3$

is a fixed recipe: 1 mole $N_2$ demands exactly 3 moles $H_2$. If you supply $H_2$ and $N_2$ in any ratio other than $3:1$, one of them will "run out" before the other. Molecules react in whole-number ratios (Dalton), so the reaction stops the instant one species hits zero — the rest just sits there.

HOW to solve — the universal 4-step method

The steps:

1. Balance the equation (coefficients $\nu_i$ are non-negotiable).
2. Convert every given mass/volume to moles: $n = m/M$.
3. Compute $R_i = n_i/\nu_i$ for each reactant → smallest is limiting.
4. Scale from the limiting reagent to get product (and leftover excess).

Worked Example 1 — the classic ammonia synthesis

Q: $28\text{ g}$ $N_2$ reacts with $10\text{ g}$ $H_2$ to form $NH_3$. Find limiting reagent, mass of $NH_3$, and leftover excess.

$N_2 + 3H_2 \rightarrow 2NH_3$

Step 1 — moles. $n_{N_2} = \frac{28}{28} = 1\text{ mol}, \qquad n_{H_2} = \frac{10}{2} = 5\text{ mol}$ Why this step? Recipe counts particles; must move from grams to moles first.

Step 2 — ratios $R_i$. $R_{N_2} = \frac{1}{1} = 1, \qquad R_{H_2} = \frac{5}{3} = 1.67$ Why this step? Whoever funds fewer "reaction sets" runs out first. $1 < 1.67$, so ==$N_2$ is limiting==.

Step 3 — product from limiting reagent. $n_{NH_3} = n_{N_2}\times\frac{2}{1} = 1\times 2 = 2\text{ mol} \Rightarrow m = 2\times 17 = 34\text{ g}$ Why this step? Only the limiting reagent decides yield; use its mole-ratio to $NH_3$.

Step 4 — leftover $H_2$. $H_2 \text{ used} = 1\times 3 = 3\text{ mol}; \quad \text{leftover} = 5 - 3 = 2\text{ mol} = 4\text{ g}$ Why this step? Excess reagent doesn't vanish; subtract what got consumed.

Worked Example 2 — masses that "look bigger" fool you

Q: $50.0\text{ g}$ Al reacts with $80.0\text{ g}$ $O_2$. Which is limiting? ($M_{Al}=27$, $M_{O_2}=32$)

$4Al + 3O_2 \rightarrow 2Al_2O_3$

Moles: $n_{Al} = 50/27 = 1.85\text{ mol}$, $n_{O_2} = 80/32 = 2.50\text{ mol}$.

Ratios: $R_{Al} = \frac{1.85}{4} = 0.463, \qquad R_{O_2} = \frac{2.50}{3} = 0.833$ Smallest is $R_{Al}$ ⇒ Al is limiting even though $O_2$'s mass is bigger. Why this step? Big mass ≠ big supply once divided by coefficient and molar mass.

Product: $n_{Al_2O_3} = 1.85\times\frac{2}{4} = 0.926\text{ mol} \Rightarrow m = 0.926\times 102 = 94.4\text{ g}$.

Worked Example 3 — percent yield tie-in

Q: In Example 1, only $28\text{ g}$ of $NH_3$ is actually collected. Percent yield?

$\%\text{ yield} = \frac{28}{34}\times 100 = 82.4\%$ Why this step? Theoretical yield ($34\text{ g}$) came from the limiting reagent; actual is what lab gives.

Flashcards

Connections

• Balancing chemical equations — coefficients $\nu_i$ come from here.
• Mole concept and Avogadro number — why moles are the currency.
• Stoichiometric calculations — general mass↔mole↔mass conversions.
• Percent yield and purity — direct downstream application.
• Concentration and molarity — solution-phase limiting reagent problems.

Concept Map

Hinglish (regional understanding)