Stoichiometry, limiting reactant and yield

Chemistry · Basic concepts. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

Test this topic freeAll Chemistry topics

The idea

A balanced equation tells us the mole ratio in which substances react and form. Stoichiometry uses this ratio to find how much product a given amount of reactant can give. The route is always the same: change the given quantity to moles, use the mole ratio, then change the answer to mass or to volume at STP.

When two reactants are mixed, one usually runs out first. This is the limiting reactant. It decides how much product forms; the other reactant is left over (in excess).

The amount calculated from the equation is the theoretical yield. The amount really obtained in the laboratory is the actual yield, and it is nearly always smaller.

Rules to remember

Common mistake

Students choose the reactant with the smaller mass, or the smaller number of moles, as the limiting reactant. Always compare the moles of product each reactant could make, because the mole ratio matters.

Worked example

2.4 g of magnesium is burned in 3.2 g of oxygen. 2Mg + O2 → 2MgO. The actual mass of MgO obtained is 3.0 g. What is the percentage yield? (Mg = 24, O = 16)

  1. Moles: Mg = 2.4 ÷ 24 = 0.1 mol; O2 = 3.2 ÷ 32 = 0.1 mol.
  2. MgO possible: from Mg, 0.1 mol; from O2, 0.1 × 2 = 0.2 mol. Magnesium gives less, so it is the limiting reactant.
  3. Theoretical yield = 0.1 × 40 = 4.0 g of MgO.
  4. Percentage yield = (3.0 ÷ 4.0) × 100 = 75%.

Answer: 75%. Magnesium is the limiting reactant and oxygen is in excess.

Ask about this

Type what you are stuck on. The helper finds the explanation, worked examples and the notes to read, free and instant.

Read the full notes

Keep going

← The mole and Avogadro's numberFiltration, crystallisation and sublimation →Take a Chemistry test