D2.6 Solve limiting-reagent and percentage-yield problems

Ontario Grade 11 Chemistry

Quantities in Chemical Reactions

Ontario Grade 11 Chemistry — D2.6

Sometimes a reaction stops even though some reactant is left over. This is like running out of one ingredient while making a recipe: the ingredients that remain cannot make more of the finished product. In chemistry, the reactant that runs out first is called the limiting reagent. It determines the greatest possible amount of product. Percentage yield compares that calculated maximum with the amount of product actually obtained.

What you will learn

Use a balanced chemical equation to compare the amounts of reactants available.
Identify the limiting reagent and use it to calculate the theoretical yield.
Calculate percentage yield from an actual yield and a theoretical yield.

1. Review the tools: equations, moles, and ratios

A balanced chemical equation shows the relative numbers of particles that react and form. It has the same number of each kind of atom on both sides. The large numbers in front of formulas are coefficients. They give the mole ratios used in calculations. Subscripts are part of a substance’s formula, so do not change them when balancing an equation.

A mole is a counting unit for particles. In a mass-to-mass problem, convert each reactant mass to moles using its molar mass. Molar mass is the mass of one mole of a substance, usually written in grams per mole. Then use the balanced equation’s coefficients to connect the amount of reactant to the amount of product.

Keep units in each calculation. Dividing a mass in grams by a molar mass in grams per mole gives an amount in moles. Coefficient ratios then connect moles of reactant to moles of product.

n = \frac{m}{M}

Balance the equation before using its coefficients.
Use molar mass to convert between mass and amount in moles.
Use coefficients as mole ratios, not as mass ratios.

2. Find the limiting reagent and theoretical yield

The limiting reagent is the reactant used up first when the reaction follows the balanced equation. An excess reagent is a reactant that remains after the limiting reagent is used up. The theoretical yield is the greatest amount of product that could form from the starting amounts, assuming the limiting reagent is used completely.

To identify the limiting reagent, calculate how much of the same product could form from each reactant. Convert the given amount to moles if needed, then use the balanced equation’s mole ratio. The reactant that gives the smaller possible product amount is limiting. This comparison is fair because both results describe the same product.

At the particle level, the equation’s coefficients show the combinations of particles that react. If there are too few particles of one reactant for the available particles of the others, that reactant is used up first. The remaining particles cannot form more product without it.

After identifying the limiting reagent, use its amount to calculate the theoretical yield. Convert the product amount to the unit requested, such as grams, if needed.

theoretical yield = product amount calculated from the limiting reagent

Compare possible product amounts from each reactant; do not choose the reactant with the smallest mass.
The limiting reagent determines the theoretical yield.
The excess reagent does not set the maximum product amount.

3. Compare actual and theoretical yield

The actual yield is the amount of product obtained in a reaction. A problem may give this value. The theoretical yield is the calculated maximum based on the limiting reagent. Percentage yield expresses the actual yield as a percentage of the theoretical yield.

Use the same units for actual and theoretical yield before dividing. For example, convert one value if one is given in grams and the other in kilograms. When the units match, they cancel in the division, leaving a percentage.

Keep extra digits during the calculation and round the final result to match the precision of the supplied measurements. A reported percentage yield can sometimes be above the calculated theoretical amount. Use the values given in the problem rather than changing them.

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

Use actual yield in the numerator and theoretical yield in the denominator.
Match units before calculating.
Round the final result appropriately and include the percent sign.

4. A clear calculation routine

Write and balance the reaction first. Convert given masses to moles, then calculate the amount of the same product each reactant could form. Compare those product amounts to identify the limiting reagent. Use that reagent to find the theoretical yield. Use the actual yield only in the final percentage-yield calculation.

Label each number with its unit. The coefficient ratio itself has no unit, but it connects amounts in moles. If the question asks for a product mass, multiply its amount in moles by its molar mass. Keep unrounded calculator values until the final step to reduce rounding error.

Follow this order: balanced equation, moles, product comparison, theoretical yield, percentage yield.
Check that each conversion gives the units requested.
Do not confuse theoretical yield with actual yield.

Worked example

Aluminum chloride from two reactant masses

A reaction uses 5.40 g of aluminum and 14.2 g of chlorine gas. The reaction produces aluminum chloride. If 15.1 g of aluminum chloride is obtained, identify the limiting reagent, calculate the theoretical yield, and find the percentage yield.

Balance the reaction
The balanced equation gives the particle and mole ratios needed for the comparison. The state symbols show that aluminum is a solid, chlorine is a gas, and aluminum chloride is a solid.
$\mathrm{2Al(s) + 3Cl_2(g) \rightarrow 2AlCl_3(s)}$
Convert reactant masses to moles
Use molar masses of $26.98\ \mathrm{g/mol}$ for aluminum and $70.90\ \mathrm{g/mol}$ for chlorine gas. Divide each mass by its molar mass to find the amount available in moles.
$n(\mathrm{Al}) = \frac{5.40\ \mathrm{g}}{26.98\ \mathrm{g/mol}} = 0.200\ \mathrm{mol};\quad n(\mathrm{Cl_2}) = \frac{14.2\ \mathrm{g}}{70.90\ \mathrm{g/mol}} = 0.200\ \mathrm{mol}$
Compare possible product amounts
Calculate aluminum chloride from each reactant separately. The balanced equation shows that two moles of aluminum make two moles of product, while three moles of chlorine gas make two moles of product. The smaller product amount identifies the limiting reagent.
$0.200\ \mathrm{mol\ Al} \times \frac{2\ \mathrm{mol\ AlCl_3}}{2\ \mathrm{mol\ Al}} = 0.200\ \mathrm{mol\ AlCl_3};\quad 0.200282\ \mathrm{mol\ Cl_2} \times \frac{2\ \mathrm{mol\ AlCl_3}}{3\ \mathrm{mol\ Cl_2}} = 0.13352\ \mathrm{mol\ AlCl_3}$
Find theoretical yield
Chlorine gas gives the smaller possible product amount, so it is the limiting reagent. Use its unrounded product amount and the molar mass of aluminum chloride, $133.33\ \mathrm{g/mol}$ . The final yield is rounded to three significant figures, matching the given measurements.
$0.13352\ \mathrm{mol\ AlCl_3} \times 133.33\ \mathrm{g/mol} = 17.80\ \mathrm{g\ AlCl_3} \approx 17.8\ \mathrm{g\ AlCl_3}$
Calculate percentage yield
The actual yield and theoretical yield are both in grams, so their units cancel in the comparison. Round the result to three significant figures.
$\frac{15.1\ \mathrm{g}}{17.8\ \mathrm{g}} \times 100\% = 84.8\%$

Answer: Chlorine gas is the limiting reagent. The theoretical yield is

17.8\ \mathrm{g}

of aluminum chloride, and the percentage yield is 84.8%.

Check: The chlorine-based product amount is smaller than the aluminum-based amount, so chlorine is limiting. The actual yield is below the theoretical yield, giving a percentage below 100%.

Common mistakes and how to avoid them

Choosing the reactant with the smaller mass as the limiting reagent.

Correction: Convert reactants to moles and compare how much product each could form. Mass alone does not account for molar mass or the equation’s coefficients.

Using the excess reagent to calculate theoretical yield.

Correction: Use the limiting reagent. It is used up first and sets the maximum possible product amount.

Reversing actual yield and theoretical yield in the percentage calculation.

Correction: Place actual yield over theoretical yield, then multiply by 100%.

Changing subscripts while balancing an equation.

Correction: Change coefficients only. Changing a subscript changes the substance and invalidates the reaction.

Rounding each intermediate value too early.

Correction: Keep extra calculator digits during the steps and round the final result to an appropriate number of significant figures.

Lesson summary

Balance the reaction and convert reactant amounts to moles when needed.
Compare the product each reactant could form. The smaller amount points to the limiting reagent.
Calculate theoretical yield from the limiting reagent.
Divide actual yield by theoretical yield and multiply by 100% to find percentage yield.

Check your understanding

Question 1

For the reaction

\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}

, a mixture has

2.0\ \mathrm{mol}

of nitrogen and

5.0\ \mathrm{mol}

of hydrogen. Which reactant is limiting?

Nitrogen, because it has fewer moles.
Hydrogen, because $2.0\ \mathrm{mol}$ of nitrogen would require $6.0\ \mathrm{mol}$ of hydrogen.
Neither; the amounts are in the exact ratio.
Nitrogen, because its coefficient is smaller.

Show answer and explanation

Hydrogen, because

2.0\ \mathrm{mol}

of nitrogen would require

6.0\ \mathrm{mol}

of hydrogen.

The equation requires three moles of hydrogen for every mole of nitrogen. The available

5.0\ \mathrm{mol}

of hydrogen is less than the

6.0\ \mathrm{mol}

needed for all the nitrogen, so hydrogen is limiting.

Question 2

A product has a theoretical yield of

24.0\ \mathrm{g}

and an actual yield of

18.0\ \mathrm{g}

. What is the percentage yield?

75.0%
133%
6.00%
42.0%

Show answer and explanation

75.0%

Divide actual yield by theoretical yield and multiply by 100%:

18.0\ \mathrm{g} \div 24.0\ \mathrm{g} \times 100\% = 75.0\%

.

Question 3

Which quantity sets the theoretical yield in a limiting-reagent problem?

The mass of the excess reagent.
The amount of product actually collected.
The amount of the limiting reagent.
The total mass of all reactants.

Show answer and explanation

The amount of the limiting reagent.

The limiting reagent is used up first, so its starting amount determines the maximum amount of product that can form.

Key terms

Limiting reagent: The reactant used up first, which determines the maximum amount of product.
Excess reagent: A reactant that remains after the limiting reagent is used up.
Theoretical yield: The calculated maximum amount of product based on the limiting reagent.
Actual yield: The amount of product obtained in a reaction.
Percentage yield: The actual yield expressed as a percentage of the theoretical yield.
Molar mass: The mass of one mole of a substance, usually expressed in grams per mole.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Chemistry (SCH3U), expectation D2.6. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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D2.6 · Solve limiting-reagent and percentage-yield problems

Ontario Grade 11 Chemistry

Ontario Grade 11 Chemistry — D2.6

What you will learn

1. Review the tools: equations, moles, and ratios

2. Find the limiting reagent and theoretical yield

3. Compare actual and theoretical yield

4. A clear calculation routine

Aluminum chloride from two reactant masses

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SCH3U

About this lesson and its review