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How to calculate excess reactant remaining

To find the excess reactant remaining, calculate how much of it the limiting reactant consumes, then subtract that amount from its starting amount. Subtract quantities of the same substance in the same units.

Excess remaining = excess initially present − excess consumed. The examples below assume the stated reaction goes to completion, with no competing reactions. They are original classroom calculations, not laboratory instructions.

The method: identify, convert, subtract

  1. Write the balanced equation. Coefficients give mole ratios.
  2. Convert each starting mass to moles.
  3. Identify the limiting reactant by calculating how much of the same product each reactant could form. The smaller possible product amount sets the limit.
  4. Use the limiting reactant's moles and the coefficient ratio to calculate moles of excess reactant consumed.
  5. Subtract consumed from initial moles of the excess reactant. Multiply the remainder by its molar mass if the question asks for grams.

You can instead convert the consumed amount to grams before subtracting. Both routes work if your units match. See stoichiometry practice for help with mole ratios.

The limiting-reactant principle is explained in OpenStax Chemistry 2e, Reaction Yields. Here we extend the calculation to track the material left over.

Example 1: starting amounts in moles

For N₂ + 3H₂ → 2NH₃, begin with 1.40 mol N₂ and 3.00 mol H₂. How much nitrogen remains?

Nitrogen could form 1.40 × 2 = 2.80 mol NH₃. Hydrogen could form 3.00 × 2/3 = 2.00 mol NH₃. Hydrogen is limiting.

Nitrogen consumed = 3.00 mol H₂ × (1 mol N₂ / 3 mol H₂) = 1.00 mol N₂. Therefore nitrogen remaining = 1.40 − 1.00 = 0.40 mol N₂.

Reaction bookkeeping, in moles
QuantityN₂H₂NH₃
Initial1.403.000
Change−1.00−3.00+2.00
Final0.4002.00

The changes follow the 1:3:2 coefficients. Adding initial moles or subtracting hydrogen moles from nitrogen moles would not give the leftover nitrogen.

Example 2: grams of excess reactant left over

For 2Mg + O₂ → 2MgO, start with 7.20 g Mg and 8.00 g O₂. Use the supplied rounded molar masses: Mg = 24.0 g/mol, O₂ = 32.0 g/mol and MgO = 40.0 g/mol.

  1. Mg: 7.20 / 24.0 = 0.300 mol. O₂: 8.00 / 32.0 = 0.250 mol.
  2. Mg could form 0.300 mol MgO. O₂ could form 0.250 × 2 = 0.500 mol MgO. Mg is limiting.
  3. O₂ consumed = 0.300 × 1/2 = 0.150 mol.
  4. O₂ remaining = 0.250 − 0.150 = 0.100 mol.
  5. Mass remaining = 0.100 × 32.0 = 3.20 g O₂.

Mass check: MgO formed = 0.300 × 40.0 = 12.0 g. Product plus unused oxygen = 12.0 + 3.20 = 15.2 g, matching 7.20 + 8.00 = 15.2 g initially. Moles of molecules need not be conserved, but total mass is conserved in this closed-system calculation.

The smaller mole amount was oxygen, yet magnesium was limiting. Comparing starting moles without the coefficients would give the wrong answer.

Three checks before accepting your answer

  • Negative remainder: recheck the limiting reactant, ratio direction and units. A negative leftover amount is impossible in this model.
  • Remainder exceeds the starting amount: you may have added the amount consumed or used the product's molar mass.
  • Both reactants finish at zero: this is possible when the starting amounts exactly match the balanced ratio. There is then no excess remaining.

A reported percent yield does not, by itself, tell you the actual amount of reactant left. Low recovered yield might reflect lost product rather than incomplete reaction. Unless conversion information is given, use the complete-reaction assumption for theoretical leftovers. Practice this distinction in percent yield problems.

Six practice questions with explained answers

Assume complete reaction. Attempt each question before opening its answer.

1. N₂ + 3H₂ → 2NH₃. Start with 2.50 mol N₂ and 6.00 mol H₂. Find the excess remaining.

Answer 1

H₂ is limiting: it consumes 6.00/3 = 2.00 mol N₂. Nitrogen remaining = 2.50 − 2.00 = 0.50 mol.

2. Use the same equation with 0.500 mol N₂ and 2.10 mol H₂.

Answer 2

N₂ is limiting. Hydrogen consumed = 0.500 × 3 = 1.50 mol; remaining = 2.10 − 1.50 = 0.60 mol H₂.

3. 2Mg + O₂ → 2MgO. Start with 9.60 g Mg and 9.60 g O₂. Use Mg = 24.0 and O₂ = 32.0 g/mol.

Answer 3

Initial amounts are 0.400 mol Mg and 0.300 mol O₂. Mg consumes 0.200 mol O₂, leaving 0.100 mol = 3.20 g O₂. Equal starting masses do not imply a stoichiometric mixture.

4. 2H₂ + O₂ → 2H₂O. Start with 0.800 mol H₂ and 0.400 mol O₂. What remains?

Answer 4

Neither reactant remains: the initial 2:1 ratio matches the equation. All 0.800 mol H₂ and 0.400 mol O₂ are consumed in this idealized calculation.

5. In example 2, a student subtracts 7.20 g Mg from 8.00 g O₂ and reports 0.80 g O₂ left. Repair the first wrong step.

Answer 5

The subtraction mixes different substances. Convert 7.20 g Mg to 0.300 mol Mg, then use the 1:2 oxygen-to-magnesium ratio. Consumed oxygen is 0.150 × 32.0 = 4.80 g. The correct subtraction is 8.00 − 4.80 = 3.20 g O₂.

6. Example 2 produces only 10.8 g of recovered MgO. Does that prove more than 3.20 g O₂ remains?

Answer 6

No. Some product could have been lost during recovery even if the reaction consumed all the magnesium. Recovered product mass alone does not distinguish incomplete conversion from recovery losses.

Build a short review set

Make three cards: how to identify the limiting reactant, how to find the excess consumed, and why actual yield cannot always reveal leftover reactants. Add a fourth card with your own numbers and check its mass balance. Use the free manual flashcard maker to write these cards, or review the broader stoichiometry lesson.

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