Chemistry
Stoichiometry: 7 Solved Problems, Step by Step
Stoichiometry turns a balanced equation into a mole ratio linking grams, moles and gas volume. Learn the 4-step method with 7 solved problems and tables.
Recommended for: Grade 11 · Grade 12
Stoichiometry is the part of chemistry that uses a balanced equation to calculate how much of a substance reacts or forms. The key idea is the mole ratio: the coefficients of the equation tell you how many moles of one substance correspond to how many moles of another. Grams, liters and particles are only converted into moles first and out of moles afterwards.
The 4-step method for every stoichiometry problem
Every problem, however long, follows the same route. First write and balance the equation. Second, convert the data you are given into moles. Third, apply the mole ratio read from the coefficients. Fourth, convert the moles you found into the unit the question asks for. If you are not yet confident with step two, review moles and molar mass first, and if the equation itself is the obstacle, practise balancing chemical equations.
Conversion map: which operation for which jump
| You have | You want | Operation | Formula |
|---|---|---|---|
| Grams of A | Moles of A | divide by molar mass | n = m ÷ M |
| Moles of A | Moles of B | multiply by mole ratio | n(B) = n(A) × coeff(B) ÷ coeff(A) |
| Moles of B | Grams of B | multiply by molar mass | m = n × M |
| Moles of gas | Volume at STP | multiply by 22.4 L/mol | V = n × 22.4 |
| Moles | Particles | multiply by Avogadro’s number | N = n × 6.022 × 10²³ |
Reading the mole ratio from an equation
Take N₂ + 3 H₂ → 2 NH₃. The coefficients are 1, 3 and 2, and every ratio you may need comes from them.
| Pair | Mole ratio | Meaning |
|---|---|---|
| N₂ and H₂ | 1 : 3 | 1 mol of N₂ reacts with 3 mol of H₂ |
| N₂ and NH₃ | 1 : 2 | 1 mol of N₂ gives 2 mol of NH₃ |
| H₂ and NH₃ | 3 : 2 | 3 mol of H₂ give 2 mol of NH₃ |
Limiting reagent and percent yield
Real reactions rarely start with the exact amounts the equation asks for. One reagent runs out first, the limiting reagent, and it decides how much product forms; the other reagent is left over, in excess. On top of that, a real process loses some product, so the actual yield is lower than the theoretical yield you calculate. The percent yield compares the two.
| Concept | How to find it |
|---|---|
| Limiting reagent | moles ÷ coefficient for each reagent, the smallest value wins |
| Theoretical yield | product moles from the limiting reagent, converted to grams |
| Excess left over | moles supplied − moles consumed |
| Percent yield | actual ÷ theoretical × 100 |
The most common mistakes
| Mistake | Why it is wrong | Fix |
|---|---|---|
| Using an unbalanced equation | the ratio is not the real one | balance before any calculation |
| Comparing grams instead of moles | 10 g of two substances are different numbers of particles | convert to moles first |
| Mixing up the ratio (A/B instead of B/A) | the answer comes out inverted | put the wanted substance on top |
| Using the mass of a reagent in excess | only the limiting reagent sets the product | find the limiting reagent first |
| Percent yield above 100 % | actual and theoretical swapped | actual always goes on top |
| Forgetting the molar volume only works for gases at STP | liquids and solids have no 22.4 L/mol | use it for gases only |
The solved problems below climb from a single mole ratio to limiting reagent and percent yield. If percentages still slow you down, a quick look at percentages will make the last exercise easier. Every number has been recalculated step by step, with significant figures respected.
Solved exercises
1. Nitrogen and hydrogen react according to N₂ + 3 H₂ → 2 NH₃. How many moles of ammonia form from 6.0 mol of H₂, with N₂ in excess? base
Show solution
- The equation is already balanced. Read the coefficients: 3 mol of H₂ give 2 mol of NH₃.
- Write the mole ratio you need: 2 mol NH₃ / 3 mol H₂.
- Multiply the moles you have by that ratio: n(NH₃) = 6.0 mol H₂ × (2 / 3).
- 6.0 × 2 = 12 and 12 ÷ 3 = 4.0.
- The moles of H₂ cancel out, leaving moles of NH₃, so the unit check works.
Answer: 4.0 mol of NH₃
2. Magnesium burns in oxygen: 2 Mg + O₂ → 2 MgO. What mass of MgO forms when 12.15 g of Mg burns completely? (Mg = 24.31 g/mol ; O = 16.00 g/mol) base
Show solution
- Grams of Mg to moles of Mg: n = 12.15 g ÷ 24.31 g/mol = 0.4998 mol.
- Mole ratio from the equation: 2 mol Mg → 2 mol MgO, which is 1 : 1, so n(MgO) = 0.4998 mol.
- Molar mass of MgO: 24.31 + 16.00 = 40.31 g/mol.
- Moles back to grams: m = 0.4998 mol × 40.31 g/mol = 20.15 g.
Answer: 20.15 g of MgO
3. A student writes H₂ + O₂ → H₂O and concludes that 32.0 g of O₂ produce 18.0 g of water (H₂ in excess). Find the mistake and compute the correct mass of water. (H = 1.008 ; O = 16.00) intermedio
Show solution
- Check the atoms: the left side has 2 oxygen atoms, the right side only 1. The equation is not balanced, so its 1 : 1 ratio is meaningless.
- Balance it: 2 H₂ + O₂ → 2 H₂O. Now 1 mol O₂ gives 2 mol H₂O.
- Moles of oxygen: n(O₂) = 32.0 g ÷ 32.00 g/mol = 1.00 mol.
- Moles of water: n(H₂O) = 1.00 × (2 / 1) = 2.00 mol.
- Molar mass of water: 2 × 1.008 + 16.00 = 18.02 g/mol.
- Mass of water: m = 2.00 mol × 18.02 g/mol = 36.0 g. The student's 18.0 g is exactly half, because the missing coefficient 2 was ignored.
Answer: 36.0 g of water, twice the student's value: the wrong 1 : 1 ratio came from the unbalanced equation
4. Methane burns: CH₄ + 2 O₂ → CO₂ + 2 H₂O. How many grams of O₂ are needed to burn 8.0 g of CH₄? (C = 12.01 ; H = 1.008 ; O = 16.00) intermedio
Show solution
- Molar mass of CH₄: 12.01 + 4 × 1.008 = 16.04 g/mol.
- Moles of methane: n(CH₄) = 8.0 ÷ 16.04 = 0.4987 mol.
- Mole ratio: 1 mol CH₄ needs 2 mol O₂, so n(O₂) = 0.4987 × 2 = 0.9974 mol.
- Molar mass of O₂: 2 × 16.00 = 32.00 g/mol.
- Mass of oxygen: m = 0.9974 × 32.00 = 31.92 g.
- The data have 2 significant figures, so the final answer is rounded to 32 g.
Answer: About 32 g of O₂ (31.9 g before rounding)
5. Limestone (pure CaCO₃) reacts with excess hydrochloric acid: CaCO₃ + 2 HCl → CaCl₂ + H₂O + CO₂. What volume of CO₂ forms from 30.0 g of CaCO₃ at STP (0 °C, 1 atm, molar volume 22.4 L/mol)? (Ca = 40.08 ; C = 12.01 ; O = 16.00) intermedio
Show solution
- Molar mass of CaCO₃: 40.08 + 12.01 + 3 × 16.00 = 100.09 g/mol.
- Moles of limestone: n = 30.0 ÷ 100.09 = 0.29973 mol.
- Mole ratio: 1 mol CaCO₃ → 1 mol CO₂, so n(CO₂) = 0.29973 mol.
- At STP one mole of an ideal gas occupies 22.4 L: V = 0.29973 mol × 22.4 L/mol = 6.714 L.
- Round to 3 significant figures: V ≈ 6.71 L.
Answer: V ≈ 6.71 L of CO₂ at STP
6. N₂ + 3 H₂ → 2 NH₃. A reactor receives 28.0 g of N₂ and 8.0 g of H₂. Identify the limiting reagent, find the mass of NH₃ formed and the mass of the reagent left over. (N = 14.01 ; H = 1.008) avanzato
Show solution
- Convert both reagents to moles: n(N₂) = 28.0 ÷ 28.02 = 0.9993 mol ; n(H₂) = 8.0 ÷ 2.016 = 3.968 mol.
- Test one reagent: 0.9993 mol N₂ would need 3 × 0.9993 = 2.998 mol H₂.
- We have 3.968 mol H₂, more than 2.998 mol, so H₂ is in excess and N₂ is the limiting reagent.
- The limiting reagent decides the product: n(NH₃) = 0.9993 × 2 = 1.9986 mol.
- Molar mass of NH₃: 14.01 + 3 × 1.008 = 17.03 g/mol, so m(NH₃) = 1.9986 × 17.03 = 34.0 g.
- Leftover H₂: 3.968 − 2.998 = 0.970 mol, i.e. 0.970 × 2.016 = 1.96 g.
- Mass check: 28.0 + 8.0 = 36.0 g before and 34.04 + 1.96 = 36.00 g after, so mass is conserved.
Answer: N₂ is the limiting reagent. It gives 34.0 g of NH₃ and leaves 1.96 g of unreacted H₂
7. In a blast furnace Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂. From 200.0 g of Fe₂O₃ the furnace actually produces 126.0 g of iron. What is the percent yield? (Fe = 55.85 ; O = 16.00) avanzato
Show solution
- Molar mass of Fe₂O₃: 2 × 55.85 + 3 × 16.00 = 159.70 g/mol.
- Moles of oxide: n = 200.0 ÷ 159.70 = 1.2523 mol.
- Mole ratio: 1 mol Fe₂O₃ → 2 mol Fe, so the theoretical amount is n(Fe) = 2.5047 mol.
- Theoretical mass: 2.5047 × 55.85 = 139.9 g. This is the maximum possible, called the theoretical yield.
- Percent yield = actual ÷ theoretical × 100 = 126.0 ÷ 139.9 × 100 = 90.1 %.
- A percent yield can never exceed 100 %: if you get more, you swapped the two masses.
Answer: Percent yield ≈ 90.1 % (theoretical yield 139.9 g)
FAQ
How do you solve a stoichiometry problem step by step?
Follow four steps: 1) write and balance the equation; 2) convert the given quantity to moles (n = m ÷ M); 3) multiply by the mole ratio from the coefficients; 4) convert the moles of the wanted substance to grams, liters or particles. Moles are always the middle step.
What is a mole ratio in stoichiometry?
It is the fraction built from the coefficients of two substances in a balanced equation. In N₂ + 3 H₂ → 2 NH₃, the ratio between H₂ and NH₃ is 2 mol NH₃ / 3 mol H₂. It only works with moles, never directly with grams.
Why do you have to balance the equation before doing stoichiometry?
Because the coefficients are the mole ratios. An unbalanced equation gives the wrong ratio and every later number is off, often by a whole factor such as 2. Balance first, calculate second.
How do you find the limiting reagent?
Convert each reagent to moles, divide each by its coefficient, and compare. The smallest value belongs to the limiting reagent. It runs out first and fixes how much product can form.
How do you calculate percent yield?
Percent yield = (actual yield ÷ theoretical yield) × 100. The theoretical yield is the mass you calculate from the balanced equation; the actual yield is what you measure in the lab. The result is always 100 % or less.