Reacting Masses and Limiting Reactants

1Why this matters

A lime producer who heats limestone wants to know how much quicklime one tonne will give. A fertiliser plant must know how much hydrogen to supply for every tonne of ammonia. Mixing reactants in the wrong proportions wastes money, because whatever is in excess is left over unused.

A balanced equation is a recipe written in moles. By the end of this lesson you will be able to use it to calculate the masses of reactants and products, identify the limiting reactant, and calculate a percentage yield.

2An equation is a mole ratio

2H₂(g) + O₂(g) → 2H₂O(l)

This equation means that 2 mol of hydrogen react with 1 mol of oxygen to form 2 mol of water. The coefficients give a ratio of moles, never of grams: 2 g of hydrogen does not react with 1 g of oxygen.

3Three steps for reacting masses

1. Mass → moles of the substance you know: n = m ÷ M.
2. Moles → moles of the substance you want, using the ratio from the balanced equation.
3. Moles → mass: m = n × M.

Worked example 1: quicklime from limestone

Solution. (1) n(CaCO₃) = 50 ÷ 100 = 0.50 mol. (2) The ratio CaCO₃ : CaO is 1 : 1, so 0.50 mol of CaO forms. (3) m = 0.50 × 56 = 28 g.

The same ratio works for any mass unit: 100 tonnes of CaCO₃ gives 56 tonnes of CaO, so 1.00 tonne gives 0.560 tonnes of quicklime.

Worked example 2: magnesium oxide

Solution. n(Mg) = 6.0 ÷ 24 = 0.25 mol. Ratio Mg : MgO = 2 : 2 = 1 : 1, so 0.25 mol of MgO. m = 0.25 × 40 = 10 g.

Worked example 3: a ratio that is not 1 : 1

Solution. n(NH₃) = 34 ÷ 17 = 2.0 mol. Ratio H₂ : NH₃ = 3 : 2, so n(H₂) = 2.0 × 3 ÷ 2 = 3.0 mol. m = 3.0 × 2 = 6.0 g.

4Limiting and excess reactants

A brochette needs 1 stick and 4 pieces of meat. With 10 sticks and 32 pieces of meat, you can make only 8 brochettes (32 ÷ 4), and 2 sticks are left over. The meat limits the number you can make; the sticks are in excess.

In a reaction, the limiting reactant is the one that is completely used up. It decides how much product forms. The other reactant is in excess: some of it is left over.

To find the limiting reactant: work out the moles of each reactant, and divide each by its coefficient in the equation. The smaller result is the limiting reactant.

Worked example 4: zinc and hydrochloric acid

6.5 g of zinc is added to a solution containing 0.30 mol of hydrochloric acid. Which reactant is limiting, and what mass of hydrogen forms?

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

Solution. n(Zn) = 6.5 ÷ 65 = 0.10 mol; 0.10 ÷ 1 = 0.10. n(HCl) = 0.30 mol; 0.30 ÷ 2 = 0.15. The smaller value is zinc's, so zinc is limiting. It uses 2 × 0.10 = 0.20 mol of HCl, leaving 0.10 mol of HCl in excess. Hydrogen formed = 0.10 mol (1 : 1 with zinc), so m = 0.10 × 2 = 0.20 g.

Worked example 5: never compare grams

4 g of hydrogen is mixed with 48 g of oxygen and ignited. Which is in excess, by how much, and what mass of water forms?

Solution. n(H₂) = 4 ÷ 2 = 2.0 mol; 2.0 ÷ 2 = 1.0. n(O₂) = 48 ÷ 32 = 1.5 mol; 1.5 ÷ 1 = 1.5. Hydrogen gives the smaller value, so hydrogen is limiting, even though its mass is much smaller. 2.0 mol of H₂ needs 1.0 mol of O₂, so 0.5 mol (16 g) of oxygen is left over. Water formed = 2.0 mol = 2.0 × 18 = 36 g.

5Percentage yield

In practice, some product is always lost: some reactant does not react, or product sticks to the apparatus. The theoretical yield is the mass calculated from the equation; the actual yield is the mass obtained.

percentage yield = actual yield ÷ theoretical yield × 100

Example: heating 50 g of CaCO₃ should give 28 g of CaO. If 25.2 g is obtained, the yield is 25.2 ÷ 28 × 100 = 90.0%.

Try it: open Reactants, Products and Leftovers (version 1.3.3). In the sandwich screen, change the amounts of bread and cheese and see which runs out first; then do the same with molecules in the Molecules screen.

Practice

Use A_r: H 1, C 12, O 16, Mg 24, Cl 35.5, Ca 40, Fe 56. Try each item before you read the solution.