Limiting Reactants and Percent Yield
A food truck makes turkey sandwiches. Each sandwich needs 2 slices of bread and 3 slices of turkey. The cook has 10 slices of bread and 9 slices of turkey. How many sandwiches can she make, and what will she have left over? Chemical reactions work the same way, and by the end of this page you will use the same thinking to predict how much product a reaction can make.
Limiting and excess reactants
Let's check the cook's supplies one ingredient at a time.
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Bread: $10 \div 2 = 5$ sandwiches.
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Turkey: $9 \div 3 = 3$ sandwiches.
She can make only 3 sandwiches. Then the turkey is gone, and she is stuck with 4 slices of bread.
In the food truck, the turkey is the limiting reactant and the bread is in excess. The balanced equation plays the role of the sandwich recipe. Its coefficients tell you how many particles (or moles) of each reactant are used up together.
In a lab, you almost never have the exact amounts the equation calls for. So you have to find out which reactant runs out first.
A food truck sells sandwiches made with 2 slices of bread and 3 slices of turkey each. Use this idea to explain, in your own words, what a limiting reactant* is in a…
A food truck sells sandwiches made with 2 slices of bread and 3 slices of turkey each. Use this idea to explain, in your own words, what a limiting reactant is in a chemical reaction. Then say what an excess reactant is, and state what happens to it when the reaction stops.
Finding the limiting reactant
For a reaction, work in moles. The coefficients of a balanced equation are mole ratios, not mass ratios.
Why does this work? Dividing by the coefficient tells you how many "recipe batches" each reactant can supply. It is the same as the bread and turkey calculation above. The reactant that supplies the fewest batches runs out first.
Try it with $\text{N}_2(g) + 3\,\text{H}_2(g) \rightarrow 2\,\text{NH}_3(g)$ in a different amount: 1.0 mol of $\text{N}_2$ and 4.5 mol of $\text{H}_2$.
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$\text{N}_2$: $1.0 \div 1 = 1.0$ batch.
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$\text{H}_2$: $4.5 \div 3 = 1.5$ batches.
$\text{N}_2$ has the smaller number, so it is the limiting reactant.
Consider the balanced equation $\text{N}2(g) + 3\,\text{H}2(g) \rightarrow 2\,\text{NH}3(g)$. A reaction vessel contains 2.0 mol of $\text{N}2$ and 3.0 mol of…
Consider the balanced equation $\text{N}_2(g) + 3\,\text{H}_2(g) \rightarrow 2\,\text{NH}_3(g)$. A reaction vessel contains 2.0 mol of $\text{N}_2$ and 3.0 mol of $\text{H}_2$. Identify which reactant is limiting, and show the mole-ratio comparison that supports your answer.
Theoretical yield and leftover reactant
Once you know the limiting reactant, it controls everything. Use it, never the excess reactant, to calculate how much product forms.
You can follow the same path every time: grams of limiting reactant → moles → moles of product (using the mole ratio) → grams of product. For leftover mass, find how much of the excess reactant was used, then subtract it from what you started with.
Mateo burns magnesium in oxygen: $2\,\text{Mg}(s) + \text{O}2(g) \rightarrow 2\,\text{MgO}(s)$. He starts with 6.07 g of Mg and 4.00 g of $\text{O}2$. The molar masses…
Mateo burns magnesium in oxygen: $2\,\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\,\text{MgO}(s)$. He starts with 6.07 g of Mg and 4.00 g of $\text{O}_2$. The molar masses are Mg = 24.31 g/mol, $\text{O}_2$ = 32.00 g/mol, and MgO = 40.31 g/mol. Determine the limiting reactant and calculate the theoretical yield of MgO in grams.
Chloe reacts 5.00 g of $\text{CH}4$ with 5.00 g of $\text{O}2$ in the combustion $\text{CH}4(g) + 2\,\text{O}2(g) \rightarrow \text{CO}2(g) + 2\,\text{H}2\text{O}(g)$.…
Chloe reacts 5.00 g of $\text{CH}_4$ with 5.00 g of $\text{O}_2$ in the combustion $\text{CH}_4(g) + 2\,\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\,\text{H}_2\text{O}(g)$. The molar masses are $\text{CH}_4$ = 16.04 g/mol and $\text{O}_2$ = 32.00 g/mol. Determine which reactant is limiting, and calculate how many grams of the excess reactant remain unused when the reaction is complete.
Marcus studies the reaction $\text{Fe}2\text{O}3(s) + 3\,\text{CO}(g) \rightarrow 2\,\text{Fe}(s) + 3\,\text{CO}2(g)$. He tests three trials, each with a different mass…
Marcus studies the reaction $\text{Fe}_2\text{O}_3(s) + 3\,\text{CO}(g) \rightarrow 2\,\text{Fe}(s) + 3\,\text{CO}_2(g)$. He tests three trials, each with a different mass of reactants:
| Trial | Mass of $\text{Fe}_2\text{O}_3$ (g) | Mass of CO (g) |
|---|---|---|
| 1 | 80.0 | 42.0 |
| 2 | 80.0 | 21.0 |
| 3 | 40.0 | 42.0 |
The molar masses are $\text{Fe}_2\text{O}_3$ = 159.69 g/mol, CO = 28.01 g/mol, and Fe = 55.85 g/mol. For each trial, identify the limiting reactant and calculate the theoretical yield of iron in grams. Then explain which trial would show the greatest increase in iron produced if Marcus doubled only the mass of CO.
Percent yield
The theoretical yield is the best-case scenario. In a real lab, you almost always collect less. Chemists measure how close they got with the percent yield.
Percent yield is a score for the experiment. 100% would mean you recovered every possible gram of product. Typical lab results are lower for realistic reasons:
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Some product stays stuck to the glassware or the filter paper.
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Some reactant takes part in a side reaction that makes something else.
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The reaction is reversible and doesn't go all the way to products.
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The product is not completely dry or is not pure, which can also throw the measurement off.
In a lab, Aaliyah calculates that a reaction should produce 12.0 g of product (the theoretical yield). After she runs the experiment and dries the product, she measures…
In a lab, Aaliyah calculates that a reaction should produce 12.0 g of product (the theoretical yield). After she runs the experiment and dries the product, she measures 9.0 g. Calculate the percent yield, and state in one sentence what the percent yield tells you about the experiment.
A student mixes 10.0 g of aluminum with excess chlorine gas: $2\,\text{Al}(s) + 3\,\text{Cl}2(g) \rightarrow 2\,\text{AlCl}3(s)$. The molar masses are Al = 26.98 g/mol…
A student mixes 10.0 g of aluminum with excess chlorine gas: $2\,\text{Al}(s) + 3\,\text{Cl}_2(g) \rightarrow 2\,\text{AlCl}_3(s)$. The molar masses are Al = 26.98 g/mol and $\text{AlCl}_3$ = 133.33 g/mol. (a) Calculate the theoretical yield of $\text{AlCl}_3$ in grams. (b) The student collects 41.5 g of $\text{AlCl}_3$. Calculate the percent yield.
A chemist prepares aspirin by reacting 2.00 g of salicylic acid ($\text{C}7\text{H}6\text{O}3$, 138.12 g/mol) with 5.00 g of acetic anhydride…
A chemist prepares aspirin by reacting 2.00 g of salicylic acid ($\text{C}_7\text{H}_6\text{O}_3$, 138.12 g/mol) with 5.00 g of acetic anhydride ($\text{C}_4\text{H}_6\text{O}_3$, 102.09 g/mol) in a 1:1 mole ratio to give aspirin ($\text{C}_9\text{H}_8\text{O}_4$, 180.16 g/mol) and acetic acid. (a) Identify the limiting reactant. (b) Calculate the theoretical yield of aspirin. (c) The chemist isolates 1.85 g of aspirin. Calculate the percent yield. (d) Give two realistic reasons the actual yield is lower than the theoretical yield.
Back to the food truck: the turkey limited the cook to 3 sandwiches (theoretical yield), and she had 4 slices of bread left over (excess reactant). If she dropped one sandwich on the floor, her actual yield would be 2, and her percent yield would be $2 \div 3 \times 100\% \approx 67\%$.