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Which reactant runs out first, and how much product do you really get?

Limiting reactants and percent yield: identify the limiting and excess reactants, calculate the theoretical yield, and calculate the percent yield.

A focused Virginia SOL Chemistry answer on yield under CH.3: identifying the limiting and excess reactants, calculating the theoretical yield of product from the limiting reactant, and finding the percent yield from the actual yield.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-13

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What this topic is asking
The limiting and excess reactants
Finding the limiting reactant
Theoretical yield and percent yield
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What this topic is asking

Standard CH.3 extends stoichiometry to real reactions, where one reactant usually runs out first. Virginia expects you to identify the limiting reactant and the excess reactant, to calculate the theoretical yield of product from the limiting reactant, and to compare it with the actual yield as a percent yield. This is how stoichiometry connects to a real laboratory result.

The limiting and excess reactants

A useful analogy: making sandwiches with two slices of bread each, if you have ten slices of bread and twenty fillings, the bread limits you to five sandwiches and the fillings are in excess. In chemistry the "recipe" is the mole ratio, so you must compare moles, not masses, against that ratio.

Finding the limiting reactant

You cannot tell the limiting reactant by mass alone, because the reactants combine in a mole ratio that is rarely $1:1$ . Always work in moles and test each reactant against the balanced equation.

Theoretical yield and percent yield

The theoretical yield is the amount of product predicted by stoichiometry from the limiting reactant, assuming the reaction goes to completion with no losses. In practice the actual yield (what you measure) is usually less, because of side reactions, incomplete reactions, or losses in handling. The percent yield measures the efficiency:

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

A percent yield above $100\%$ is impossible in a clean reaction; if you calculate one, the product is impure or a measurement is wrong, because the theoretical yield is the maximum.

Limiting reactant and theoretical yield

For $\text{N}_2 + 3\,\text{H}_2 \rightarrow 2\,\text{NH}_3$ , you start with $2.0$ mol of nitrogen and $3.0$ mol of hydrogen. Find the limiting reactant and the theoretical yield of ammonia in moles.

step 1 Product from the nitrogen

Nitrogen to ammonia is $1 : 2$ , so $2.0$ mol N $_2$ could make $4.0$ mol NH $_3$ .

step 2 Product from the hydrogen

Hydrogen to ammonia is $3 : 2$ , so $3.0$ mol H $_2$ could make $3.0 \times \dfrac{2}{3} = 2.0$ mol NH $_3$ .

step 3 Choose the smaller

Hydrogen makes only $2.0$ mol of ammonia, less than nitrogen's $4.0$ mol, so hydrogen is the limiting reactant and nitrogen is in excess.

step 4 State the theoretical yield

The theoretical yield is $2.0$ mol of ammonia, set by the limiting reactant (hydrogen).

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Q1. In a reaction, $0.30$ mol of one reactant could make $0.30$ mol of product, while the other reactant could make $0.45$ mol. Which is the limiting reactant? [1 point]

Cue. The first reactant; it makes the smaller amount of product ( $0.30$ mol), so it runs out first.

Q2. A reaction has a theoretical yield of $25.0$ g and an actual yield of $20.0$ g. Calculate the percent yield. [2 points]

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

Exam-style practice questions

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

SOL (multiple choice)1 marksIn a reaction, the limiting reactant is the one that (A) is present in the largest mass (B) is completely used up first and limits the product (C) is left over at the end (D) has the largest molar mass

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The answer is (B) is completely used up first and limits the product.

The limiting reactant runs out first and therefore determines (limits) how much product can form; once it is gone, the reaction stops even if other reactants remain. The reactant left over is the excess reactant (C). Mass (A) and molar mass (D) do not by themselves decide which is limiting; you compare the available moles against the mole ratio.

The trap is choosing the reactant present in the smallest mass; the limiting reactant depends on moles and the mole ratio, not on mass alone.

SOL (tech-enhanced, fill in the blank)3 marksA student reacts magnesium with oxygen and produces

7.0

g of magnesium oxide. The theoretical yield calculated from the limiting reactant is

8.0

g. (a) Calculate the percent yield. (b) State whether the actual yield can ever exceed the theoretical yield.

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A 3-point percent-yield item.

(a) Percent yield (2 points): $\text{percent yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \dfrac{7.0}{8.0} \times 100 = 87.5\%$ .
(b) No (1 point): the actual yield cannot exceed the theoretical yield (so percent yield cannot exceed $100\%$ ), because the theoretical yield is the maximum possible from the limiting reactant; a value above $100\%$ signals impurity or measurement error.

Markers reward the percent-yield formula and the reasoning that the theoretical yield is the ceiling.

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Sources & how we know this

2018 Science Standards of Learning - Chemistry — Virginia Department of Education (2018)
Chemistry Curriculum Framework — Virginia Department of Education (2018)