For 2Al + 3Cl_2 → 2AlCl_3, calculate the moles of AlCl_3 from 0.600 mol of Al (with Cl_2 in excess). [2 points]

College Board AP 2023 (style)-style practice: Section II (long FRQ, part). Consider the reaction N_2(g) + 3H_2(g) → 2NH_3(g). A mixture contains 14.0 g of N_2 and 6.00 g of H_2. (a) Determine the limiting reactant. (b) Calculate the theoretical yield of NH_3 in grams. (c) If 11.9 g of NH_3 is actually produced, calculate the percent yield. (d) Justify why one reactant is left over.

A 4-point quantitative FRQ on stoichiometry (M(N_2) = 28.0, M(H_2) = 2.02, M(NH_3) = 17.0 g/mol). (a) Limiting reactant (1 point): n(N_2) = 14.028.0 = 0.500 mol; n(H_2) = 6.002.02 = 2.97 mol. N_2 needs 3 × 0.500 = 1.50 mol H_2, which is available, so N_2 is limiting. (b) Theoretical yield (1 point): 0.500 mol N_2 × 2\ mol NH_31\ mol N_2 = 1.00 mol NH_3; mass = 1.00 × 17.0 = 17.0 g. (c) Percent yield (1 point): 11.917.0 × 100\% = 70.0\%. (d) Justify (1 point): H_2 is in excess; only 1.50 mol of the 2.97 mol available reacts, so about 1.47 mol H_2 is left over once the limiting N_2 is used up. Markers reward identifying the limiting reactant by mole comparison, the theoretical yield from the mole ratio, the percent yield, and explaining the leftover excess reactant.

United StatesChemistrySyllabus dot point

How do the coefficients of a balanced equation let us calculate amounts of reactants and products, including the limiting reactant?

Topic 4.5 Stoichiometry: use mole ratios from a balanced equation to relate amounts of reactants and products, and determine the limiting reactant, theoretical yield and percent yield.

A focused answer to AP Chemistry Topic 4.5, covering mole ratios from balanced equations, mass-to-mass calculations, the limiting reactant, theoretical yield and percent yield, with full worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Mole ratios
The limiting reactant
Theoretical and percent yield
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What this topic is asking

The College Board (Topic 4.5) wants you to use the mole ratios in a balanced equation to relate the amounts of reactants and products, find the limiting reactant, and calculate the theoretical yield and percent yield. Stoichiometry is the quantitative heart of the course: every reaction calculation, in solution, in gases, or by mass, runs through the mole ratio of the balanced equation.

Mole ratios

This is the engine of every stoichiometry problem. To convert from an amount of one substance to an amount of another, you go through moles: convert the given quantity to moles, apply the mole ratio, then convert to whatever quantity is asked for. The general path is mass to moles (divide by molar mass), moles to moles (multiply by the ratio), moles to mass (multiply by molar mass).

The limiting reactant

A common error is to compare raw masses or even raw moles without using the mole ratio. The correct method respects the equation: $0.40$ mol of a reactant that needs twice as much of a partner requires $0.80$ mol of the partner, so if only $0.60$ mol is available, the partner is limiting. The limiting reactant sets every product amount.

Theoretical and percent yield

The theoretical yield is the maximum amount of product calculable from the limiting reactant, assuming the reaction goes to completion. In practice less is obtained, because reactions may not go to completion, side reactions occur, or product is lost in handling. The percent yield measures this:

\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%

Both yields must be in the same units (usually grams or moles). A percent yield above 100% signals an error or an impure product, since you cannot make more than the theoretical maximum.

A mass-to-mass calculation with a limiting reactant

How many grams of water form when $4.00$ g of $\text{H}_2$ reacts with $16.0$ g of $\text{O}_2$ in $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ ?

step 1 Convert to moles

$n(\text{H}_2) = \dfrac{4.00}{2.02} = 1.98$ mol; $n(\text{O}_2) = \dfrac{16.0}{32.0} = 0.500$ mol.

step 2 Find the limiting reactant

The ratio needed is $2\ \text{H}_2 : 1\ \text{O}_2$ . To use all $0.500$ mol $\text{O}_2$ would need $1.00$ mol $\text{H}_2$ , which is available; to use all $1.98$ mol $\text{H}_2$ would need $0.99$ mol $\text{O}_2$ , which is not. So $\text{O}_2$ is limiting.

step 3 Use the mole ratio

$0.500$ mol $\text{O}_2 \times \dfrac{2\ \text{mol H}_2\text{O}}{1\ \text{mol O}_2} = 1.00$ mol $\text{H}_2\text{O}$ .

step 4 Convert to mass

$m = 1.00\ \text{mol} \times 18.0\ \text{g/mol} = 18.0$ g of water.

Try this

Q1. For $2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$ , calculate the moles of $\text{AlCl}_3$ from $0.600$ mol of $\text{Al}$ (with $\text{Cl}_2$ in excess). [2 points]

Cue. $0.600 \times \dfrac{2}{2} = 0.600$ mol $\text{AlCl}_3$ (the Al:AlCl3 ratio is $1:1$ ).

Q2. A reaction has a theoretical yield of $25.0$ g but produces $20.0$ g. Calculate the percent yield. [1 point]

Cue. $\dfrac{20.0}{25.0} \times 100\% = 80.0\%$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). Consider the reaction

\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)

. A mixture contains

14.0

g of

\text{N}_2

and

6.00

g of

\text{H}_2

. (a) Determine the limiting reactant. (b) Calculate the theoretical yield of

\text{NH}_3

in grams. (c) If

11.9

g of

\text{NH}_3

is actually produced, calculate the percent yield. (d) Justify why one reactant is left over.

Show worked answer →

A 4-point quantitative FRQ on stoichiometry ( $M(\text{N}_2) = 28.0$ , $M(\text{H}_2) = 2.02$ , $M(\text{NH}_3) = 17.0$ g/mol).

(a) Limiting reactant (1 point): $n(\text{N}_2) = \dfrac{14.0}{28.0} = 0.500$ mol; $n(\text{H}_2) = \dfrac{6.00}{2.02} = 2.97$ mol. $\text{N}_2$ needs $3 \times 0.500 = 1.50$ mol $\text{H}_2$ , which is available, so $\text{N}_2$ is limiting.
(b) Theoretical yield (1 point): $0.500$ mol $\text{N}_2 \times \dfrac{2\ \text{mol NH}_3}{1\ \text{mol N}_2} = 1.00$ mol $\text{NH}_3$ ; mass $= 1.00 \times 17.0 = 17.0$ g.
(c) Percent yield (1 point): $\dfrac{11.9}{17.0} \times 100\% = 70.0\%$ .
(d) Justify (1 point): $\text{H}_2$ is in excess; only $1.50$ mol of the $2.97$ mol available reacts, so about $1.47$ mol $\text{H}_2$ is left over once the limiting $\text{N}_2$ is used up.

Markers reward identifying the limiting reactant by mole comparison, the theoretical yield from the mole ratio, the percent yield, and explaining the leftover excess reactant.

AP 2021 (style)1 marksSection I (multiple choice). In a reaction,

0.40

mol of reactant A reacts with B in a

1:2

ratio (A:B). If

0.60

mol of B is available, the limiting reactant is (A) A (B) B (C) neither, they are exactly matched (D) cannot be determined. Justify your reasoning.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

A needs twice as much B: $0.40$ mol A requires $0.80$ mol B, but only $0.60$ mol B is available, so B runs out first and is the limiting reactant. The trap is comparing raw moles without using the $1:2$ ratio.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)