AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/4/4-5)
Unit 4 · Topic 4.5
4.5 Stoichiometry
The coefficients in a balanced equation give mole ratios, which let you calculate how much product forms from a given amount of reactant, or how much reactant you need. These ratios connect to grams through molar mass, to gases through PV = nRT and to solutions through molarity.
Key terms
- mole ratio
- limiting reactant
- theoretical yield
- molarity
- ideal gas law
The mole ratio is the bridge
Coefficients count moles, not grams. In 2H₂ + O₂ → 2H₂O, 2 mol H₂ react with 1 mol O₂ to make 2 mol H₂O. You can never go directly from grams of one substance to grams of another; you always pass through moles and the mole ratio.
Every stoichiometry problem follows the same three steps.
- Convert what you're given into moles: grams ÷ molar mass, or M × L for a solution, or PV/RT for a gas.
- Use the mole ratio from the balanced equation: moles wanted = moles given × (coefficient wanted ÷ coefficient given).
- Convert moles of what you want into the units asked for.
Limiting reactant
When amounts of two reactants are given, one usually runs out first. That one is the limiting reactant, and it sets how much product can form. The other is in excess, and some is left over.
To find the limiting reactant, convert each reactant to moles, then work out how much product each one could make. The reactant that makes less product is limiting. Or calculate how much of reactant B is needed to use up all of reactant A, and compare with what you have.
The amount of product calculated from the limiting reactant is the theoretical yield. The amount actually collected in the lab is usually less, because of side reactions or losses. Percent yield = (actual ÷ theoretical) × 100.
A percent yield over 100% is a red flag that the product still contains something else, such as water from incomplete drying or an impurity. A low percent yield can come from an incomplete reaction, side reactions or product lost while transferring or filtering.
Linking to gases and solutions
For gases, use n = PV/RT to turn a volume into moles, or V = nRT/P to turn moles into a volume, with T in kelvins.
For solutions, use moles = M × V (in liters). This is how you calculate the mass of precipitate formed from a known volume of solution, or the volume of one solution needed to react with another.
Mole ratios work just as well between two solutions: to find the volume of 0.100 M NaOH needed to react with a given amount of acid, find moles of acid, use the ratio to get moles of NaOH, then divide by 0.100 mol/L.
Checks that catch errors
Total mass of reactants used equals total mass of products formed. The limiting reactant must give the smaller amount of product. And the answer should make sense: if 10 g of reactant gave 500 g of product, something went wrong.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Grams to grams
Propane burns according to C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g). What mass of CO₂ forms when 10.0 g of propane burns completely?
Show the solutionHide the solution
- Step 1: Molar mass of C₃H₈ = 3(12.01) + 8(1.008) = 44.09 g/mol. Moles = 10.0 ÷ 44.09 = 0.2268 mol.
- Step 2: Mole ratio: 3 mol CO₂ per 1 mol C₃H₈, so moles CO₂ = 0.2268 × 3 = 0.6804 mol.
- Step 3: Molar mass CO₂ = 44.01 g/mol. Mass = 0.6804 × 44.01 = 29.9 g.
Answer: 29.9 g CO₂
- Example 2Calculator allowed
Limiting reactant (classic trap)
5.00 g of aluminum reacts with 10.0 g of chlorine gas: 2Al(s) + 3Cl₂(g) → 2AlCl₃(s). Which reactant is limiting, what mass of AlCl₃ forms, and what mass of the excess reactant remains?
Show the solutionHide the solution
- Step 1: Moles: Al = 5.00 ÷ 26.98 = 0.1853 mol. Cl₂ = 10.0 ÷ 70.90 = 0.1410 mol.
- Step 2: Al needs 3/2 as many moles of Cl₂: 0.1853 × 1.5 = 0.2780 mol Cl₂. Only 0.1410 mol is available, so Cl₂ is limiting.
- Step 3: The trap: aluminum has the smaller mass, but chlorine is limiting. Compare moles adjusted by coefficients, not grams.
- Step 4: AlCl₃ formed = 0.1410 × (2/3) = 0.09403 mol. Molar mass AlCl₃ = 26.98 + 3(35.45) = 133.33 g/mol. Mass = 0.09403 × 133.33 = 12.5 g.
- Step 5: Al used = 0.09403 mol (1 : 1 with AlCl₃). Al left = 0.1853 − 0.0940 = 0.0913 mol × 26.98 = 2.46 g.
Answer: Cl₂ is limiting; 12.5 g AlCl₃ forms; 2.46 g Al remains.
- Example 3Calculator allowed
Stoichiometry with a gas and a solution
(a) What volume of H₂ gas, at 25 °C and 1.00 atm, forms when 0.500 g of Mg reacts with excess HCl? Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). (b) What mass of AgCl precipitates when 25.0 mL of 0.200 M AgNO₃ reacts with excess NaCl?
Show the solutionHide the solution
- Step 1: (a) Moles Mg = 0.500 ÷ 24.31 = 0.02057 mol. Ratio 1 : 1, so 0.02057 mol H₂.
- Step 2: V = nRT/P = (0.02057)(0.08206)(298.15) ÷ 1.00 = 0.503 L.
- Step 3: (b) Moles AgNO₃ = 0.200 M × 0.0250 L = 5.00 × 10⁻³ mol. Ag⁺ + Cl⁻ → AgCl is 1 : 1, so 5.00 × 10⁻³ mol AgCl.
- Step 4: Mass = 5.00 × 10⁻³ × 143.32 g/mol = 0.717 g.
Answer: (a) 0.503 L H₂; (b) 0.717 g AgCl
Common mistakes
- Using a mole ratio with grams. Convert to moles first.
- Picking the limiting reactant by comparing masses, or by comparing moles without the coefficients.
- Using an unbalanced equation. Balance first, every time.
- Forgetting to convert mL to L or °C to K in solution and gas problems.
On the exam
- Stoichiometry is part of most quantitative free-response questions. Lay out each conversion with units so partial credit is possible even if one number slips.
- Expect to be asked which reactant is limiting and to justify it with calculations, or to explain how an experimental error would change the calculated yield.
Connected topics
Videos
Check yourself
5 questions on 4.5 Stoichiometry. Pick an answer to see if you got it, and why.
Aluminum reacts with chlorine gas: 2Al(s) + 3Cl₂(g) → 2AlCl₃(s). If 5.40 g of Al (0.200 mol) reacts with 0.250 mol of Cl₂, what is the maximum mass of AlCl₃ (molar mass 133.3 g/mol) that can form?
A 10.0 g sample of CaCO₃ (molar mass 100.09 g/mol) reacts completely with excess HCl: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). What volume of CO₂ is produced at 298 K and 1.00 atm? (R = 0.08206 L·atm/(mol·K))
Excess NaCl(aq) is added to 50.0 mL of 0.200 M AgNO₃(aq), and all the Ag⁺ precipitates as AgCl (molar mass 143.3 g/mol). What mass of AgCl forms?
A sealed steel vessel contains 10.0 g of H₂ and 64.0 g of O₂. A spark starts the reaction below, which goes to completion.
2H₂(g) + O₂(g) → 2H₂O(g)
Molar masses: H₂, 2.016 g/mol; O₂, 32.00 g/mol; H₂O, 18.02 g/mol.
Described laboratory experiment
Which reactant is limiting, and what is the maximum mass of water that can form?
What mass of the excess reactant remains after the reaction is complete?
0 of 5 answered