The method in four steps
- Balance the equation. The coefficients count moles: 1 mol Fe₂O₃ reacts with 3 mol CO to give 2 mol Fe and 3 mol CO₂.
- Convert grams to moles with
n = m ÷ M, using the molar mass from the atomic weights. - Divide each reactant's moles by its coefficient. The smallest value is the number of times the reaction can run as written, and it belongs to the limiting reagent.
- Calculate the theoretical yield: that smallest value × the product's coefficient × the product's molar mass.
In Excel, with moles ÷ coefficient in F2:F3, =MIN(F2:F3) gives the number of reaction runs and =INDEX(A2:A3,MATCH(MIN(F2:F3),F2:F3,0)) returns the name of the limiting reagent.
Working through the example
The molar masses come from the atomic weights Fe = 55.845, C = 12.011 and O = 15.999 g/mol. Fe₂O₃ is 2 × 55.845 + 3 × 15.999 = 159.687 g/mol, CO is 28.010 g/mol. So 500 g of Fe₂O₃ is 500 ÷ 159.687 = 3.1311 mol, and 250 g of CO is 250 ÷ 28.010 = 8.9254 mol.
Dividing by the coefficients: Fe₂O₃ 3.1311 ÷ 1 = 3.1311, CO 8.9254 ÷ 3 = 2.9751. CO has the smaller value, so CO is the limiting reagent. The reaction can run 2.9751 mol times, which gives 2.9751 × 2 = 5.9502 mol of iron, or 5.9502 × 55.845 = 332.3 g. It also gives 2.9751 × 3 × 44.009 = 392.8 g of CO₂.
Worked example: limiting reagent and yield for Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂
Coefficients, masses and the atomic weight of iron are yellow inputs. Molar masses, moles, the limiting reagent and the yields are formulas.
| A | B | C | D | E | F | |
|---|---|---|---|---|---|---|
| 1 | Substance | Coefficient | Molar mass (g/mol) | Mass (g) | Moles = m ÷ M | Moles ÷ coefficient |
| 2 | Fe₂O₃ | 1 | 159.687 | 500.0 | 3.1311 | 3.1311 |
| 3 | CO | 3 | 28.010 | 250.0 | 8.9254 | 2.9751 |
| 4 | Fe (product) | 2 | 55.845 | |||
| 5 | CO₂ (product) | 3 | 44.009 | |||
| 6 | ||||||
| 7 | Limiting reagent (smallest moles ÷ coefficient) | CO | ||||
| 8 | Times the reaction runs = MIN(F2:F3), mol | 2.9751 | ||||
| 9 | Theoretical yield of Fe = F8 × 2 × M(Fe), g | 332.3 | ||||
| 10 | Actual yield of Fe, g | 301.0 | ||||
| 11 | Percent yield = actual ÷ theoretical | 90.6% | ||||
| 12 | Fe₂O₃ left over (excess reagent), g | 24.9 | ||||
| 13 | CO₂ formed (theoretical), g | 392.8 | ||||
- E2
=D2/C2Moles = mass ÷ molar mass - F2
=E2/B2Moles ÷ coefficient - F7
=INDEX(A2:A3,MATCH(MIN(F2:F3),F2:F3,0))Name of the limiting reagent - F9
=F8*B4*C4Theoretical yield of iron - F11
=F10/F9Percent yield - F12
=(F2-F8)*B2*C2Excess Fe₂O₃ left over
Atomic weights: IUPAC abridged values. The masses and the actual yield are teaching numbers. Calculations only, no lab procedure.
Why comparing raw moles or grams goes wrong
Looking at the moles alone, CO seems to be in large excess: 8.93 mol against 3.13 mol. But each Fe₂O₃ needs three CO, so 3.13 mol of Fe₂O₃ would need 9.39 mol of CO, more than the 8.93 mol available. Comparing grams is even less reliable, because molar masses differ. Always compare moles ÷ coefficient.
A second way to check: take one reactant and calculate how much of the other it needs. 8.9254 mol CO needs 8.9254 ÷ 3 = 2.9751 mol Fe₂O₃, and 3.1311 mol are available, so Fe₂O₃ is in excess and CO is limiting. Both methods always agree.
Excess reagent left over and conservation of mass
The excess reagent is what remains when the limiting reagent is used up: (moles available ÷ coefficient − reaction runs) × coefficient × molar mass. Here (3.1311 − 2.9751) × 1 × 159.687 = 24.9 g of Fe₂O₃ left over.
Mass is a good final check. The reaction used 500 − 24.9 = 475.1 g of Fe₂O₃ and all 250 g of CO, 725.1 g in total. It made 332.3 g of Fe and 392.8 g of CO₂, also 725.1 g. If the masses do not balance, a molar mass or a coefficient is wrong.
Percent yield
The theoretical yield assumes every bit of the limiting reagent reacts and nothing is lost. Real reactions fall short. Percent yield = actual yield ÷ theoretical yield. If 301.0 g of iron is recovered, the percent yield is 301.0 ÷ 332.3 = 90.6%. A percent yield above 100% usually means the product is wet or impure, not that the calculation went right.
The same steps work for any reaction: balance, convert to moles, divide by coefficients, take the minimum and convert back to grams. With the atomic weights in a lookup table, XLOOKUP and SUMPRODUCT can calculate the molar masses for you.
Common mistakes
- Comparing grams. 500 g against 250 g says nothing, because the molar masses differ.
- Comparing moles without the coefficients. CO has more moles but is still limiting, because it is needed three times over.
- Using an unbalanced equation. Without the 3 in front of CO, the calculation would pick the wrong reagent.
- Forgetting the product coefficient. Each run makes 2 mol of Fe, so multiply the runs by 2 before converting to grams.
- Rounding moles too early. Keep four significant figures or more in the intermediate steps and round the final answer.
- Taking a percent yield above 100% at face value. It points to a wet or impure product, or to a calculation error.
Frequently asked questions
How do you find the limiting reagent?
Convert the mass of each reactant to moles, divide by its coefficient in the balanced equation and compare. The reactant with the smallest result is the limiting reagent.
How do you calculate the theoretical yield?
Multiply the smallest moles ÷ coefficient value by the product's coefficient and its molar mass. In the example: 2.9751 × 2 × 55.845 = 332.3 g of iron.
Can the reactant with more moles be the limiting reagent?
Yes. In the example CO has 8.93 mol against 3.13 mol of Fe₂O₃, but the reaction needs three CO per Fe₂O₃, so CO runs out first.
How do you calculate how much excess reagent is left?
Subtract the moles used, reaction runs × coefficient, from the moles available and multiply by the molar mass. Here 24.9 g of Fe₂O₃ is left over.