Stoichiometry
Use reacting proportions to find the limiting reactant, product and remaining excess.
Before this connection: Moles and solution concentration; For the gas extension, first read States of matter and gases.
AP Chemistry Topic 4.5 · CED printed p. 82 / PDF p. 89. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Moles and molarity
- Ionic charges and molecular formulas
- Intermolecular versus intramolecular interactions
- Ideal gas equation with kelvin temperature
A balanced equation gives a bookkeeping path from a known amount to an unknown amount. Convert to moles first, then use the coefficient ratio, then convert to the requested unit. With multiple reactants, calculate how much product each could support; the smaller result sets the maximum yield. That reactant is limiting. Subtract the amount actually used from the initial excess amount. This is conservation with arithmetic, not a special trick.
For 2Al + 3Cl2 → 2AlCl3, three moles of chlorine react with two moles of aluminum. The ratio does not mean 3 g react with 2 g because different particles have different masses. A ratio such as 2 mol AlCl3/3 mol Cl2 works because it converts numbers of reacting units. If reactant amount doubles while all others remain in sufficient excess, the maximum product doubles; once another reactant becomes limiting, that prediction stops working.
For solutions, n = MV with V in liters gives the number of available moles. For gases, n = PV/RT supplies the same bridge, with temperature in kelvin and compatible units for R. Then use the reaction ratio. Gas volumes share the coefficient ratio only at the same pressure and temperature under ideal-gas behavior. If the sample is collected over water, the gas pressure must exclude water vapor when its partial pressure is supplied. Concentrations after mixing require the total solution volume, not whichever reagent volume is convenient.
Let's work through it
Mix 25.0 mL of 0.200 M CaCl2 and 40.0 mL of 0.150 M Na2CO3. Assume complete CaCO3 precipitation, additive volumes, and molar mass 100.09 g mol−1.
- n(Ca2+) = 0.0250 L × 0.200 mol L−1 = 0.00500 mol. n(CO32−) = 0.0400 × 0.150 = 0.00600 mol.
- The net ionic ratio is 1:1, so calcium is limiting; 0.00500 mol solid forms. Its mass is 0.00500 ×100.09 = 0.500 g.
- Carbonate remaining = 0.00100 mol. Divide by 0.0650 L, the total solution volume: [CO32−] = 0.0154 M. This ignores the small solubility of the solid, as specified.
A supplied gas sample contains 0.0100 mol product gas at 300 K, with product partial pressure 0.950 atm. Use R =0.08206 L atm mol⁻¹ K⁻¹.
- V =nRT/P =0.0100×0.08206×300/0.950 =0.259 L. Kelvin temperature and matching pressure units make the calculation valid.
- If pressure came from wet gas, first subtract the water-vapor contribution. The product moles then connect to reactant moles through the balanced equation.
- The solution worked example above uses n =MV instead; both are routes to chemical amount before applying the same stoichiometric logic.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
Why can’t you compare 2.0 g A and 3.0 g B to decide the limiting reactant directly?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
Why can’t you compare 2.0 g A and 3.0 g B to decide the limiting reactant directly?
Worked answer and reason
Convert each mass to moles and compare amount/coefficient. Different molar masses and required ratios can reverse the mass-based guess.
Coefficients compare chemical units, not masses.
Retrieval 2
What happens if you double the limiting reagent while keeping all others sufficiently in excess?
Worked answer and reason
The maximum product doubles. If an initially excess reagent becomes limiting, the proportional prediction no longer holds.
The reaction ratio applies to the amount that actually reacts.
Retrieval 3
Why does M×V require volume in liters when M is mol L⁻¹? How do you use those moles in a reaction?
Worked answer and reason
Liters cancel L⁻¹, leaving moles. Apply the balanced mole ratio, then divide any final dissolved amount by the final solution volume to obtain its concentration.
Dimensional cancellation provides an amount before stoichiometric comparison.
Retrieval 4
A gas is collected over water. Which pressure belongs in n =PV/RT for the dry product gas?
Worked answer and reason
Use the product partial pressure: total pressure minus supplied water-vapor pressure. Use kelvin temperature and compatible R units, then apply the reaction mole ratio.
The measured total pressure includes more than the product species.
Catch the tempting mistake
- The reactant with smaller mass must be limiting.
- Compare moles divided by stoichiometric coefficients.
- Use one reagent’s volume to find final concentration.
- Use the final solution volume, with additive volumes only when the model permits it.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 4 problem 8 · ap application · symbolic
- Unit 4 problem 9 · ap application · data
- Unit 4 problem 10 · ap synthesis · multiple
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For the teacher
- Have students justify their limiting choice before calculating grams.
- Ask how a volume or temperature error propagates through n = PV/RT.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 4.5.A
- Explain changes in the amounts of reactants and products based on the balanced reaction equation for a chemical process.
- 4.5.A.1
- Because atoms must be conserved during a chemical process, it is possible to calculate product amounts by using known reactant amounts, or to calculate reactant amounts given known product amounts.
- 4.5.A.2
- Coefficients of balanced chemical equations contain information regarding the proportionality of the amounts of substances involved in the reaction. These values can be used in chemical calculations involving the mole concept.
- 4.5.A.3
- Stoichiometric calculations can be combined with the ideal gas law and calculations involving molarity to quantitatively study gases and solutions.
Science references for this teaching
- College Board AP Chemistry Course and Exam Description, Fall 2024 — Exact unit learning requirements and exclusions; supplied PDF is pinned by the integration catalog.
- OpenStax Chemistry 2e: Reaction Stoichiometry — Scientific check: mole ratios and limiting reactants; all numerical tasks are newly authored.
Glossary and deeper reading · 5 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Mole
A mole is an amount of substance containing exactly 6.02214076 × 10²³ specified entities. The entities can be atoms, molecules, ions, or formula units, so the counted entity must be named. For example, one mole of oxygen molecules contains two moles of oxygen atoms.
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Molarity
Molarity is the amount of a solute in moles divided by the final solution volume in liters. It describes concentration rather than the total amount of solute. For example, 0.10 mole dissolved to make 0.50 liter of solution gives a molarity of 0.20 mole per liter.
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Molecular formula
A molecular formula gives the actual number of each kind of atom in a molecule. It can be a whole-number multiple of the empirical formula. For example, glucose has molecular formula C₆H₁₂O₆ and empirical formula CH₂O.
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Ideal gas law
The ideal gas law relates pressure, volume, amount, and absolute temperature in an idealized gas model. It treats particles as having negligible volume and no intermolecular attractions. For example, at fixed amount and temperature, doubling the container volume halves the ideal gas pressure.
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Partial pressure
A gas's partial pressure is the pressure it would exert alone in the mixture's volume at the same temperature. In an ideal mixture, the component partial pressures add to the total pressure. For example, a component with mole fraction 0.20 contributes 0.20 of the total ideal pressure.