Reacting Masses and Limiting Reactants

(Higher Tier only) Reacting-mass calculations, using moles to balance equations and limiting reactants are all assessed only at Higher Tier.

A balanced equation is more than a list of substances. The multipliers give the mole ratio in which the substances react, and that ratio is the key to every reacting-mass calculation.

$2\text{Mg}+{\text{O}}_2\to 2\text{MgO}$

This reads as: 2 moles of magnesium react with 1 mole of oxygen to make 2 moles of magnesium oxide. The ratio of Mg to MgO is 2 : 2, which simplifies to 1 : 1.

The multipliers in a balanced equation give the ratio of moles, not the ratio of masses. Always convert masses to moles before using the ratio.

Every reacting-mass calculation follows the same three-step route: convert the known mass to moles, use the mole ratio from the equation to find the moles of the substance you want, then convert those moles back to a mass.

The three-step route turns a mass of reactant into a mass of product. Set each step out clearly so an examiner can follow your reasoning.

Worked example — what mass of magnesium oxide is made when 48 g of magnesium burns completely?

$2\text{Mg}+{\text{O}}_2\to 2\text{MgO}$

Step 1 — moles of Mg (Ar = 24):

$\text{moles Mg}=\frac{48}{24}=2\text{mol}$

Step 2 — mole ratio. Mg : MgO is 2 : 2, so 2 mol of Mg gives 2 mol of MgO.

Step 3 — mass of MgO (Mr = 24 + 16 = 40):

$\text{mass MgO}=40\times 2=80\text{g}$

So 48 g of magnesium produces 80 g of magnesium oxide. The extra 32 g is the mass of oxygen taken from the air, consistent with conservation of mass.

Many equations have ratios such as 1 : 2 or 2 : 3. The method is identical; you just apply the correct ratio in step 2.

Worked example — what mass of aluminium oxide forms when 5.4 g of aluminium reacts fully with oxygen?

$4\text{Al}+3{\text{O}}_2\to 2{\text{Al}}_2{\text{O}}_3$

Step 1 — moles of Al (Ar = 27):

$\text{moles Al}=\frac{5.4}{27}=0.2\text{mol}$

Step 2 — mole ratio. Al : Al₂O₃ is 4 : 2, which simplifies to 2 : 1. So moles of Al₂O₃ = 0.2 ÷ 2 = 0.1 mol.

Step 3 — mass of Al₂O₃. Mr = (2 × 27) + (3 × 16) = 54 + 48 = 102.

${\text{mass Al}}_2{\text{O}}_3=102\times 0.1=10.2\text{g}$

The ratio 4 : 2 means fewer moles of product than reactant, so halving the moles at step 2 is essential.

You can work backwards: if you measure the masses that react, converting them to moles gives the balancing numbers. Convert each mass to moles, then write the moles as the simplest whole-number ratio.

Worked example — 0.6 g of magnesium combines exactly with 0.4 g of oxygen. Find the formula of the oxide.

Element	Mass (g)	Ar	Moles (mass ÷ Ar)
Mg	0.6	24	0.6 ÷ 24 = 0.025
O	0.4	16	0.4 ÷ 16 = 0.025

The ratio of moles Mg : O is 0.025 : 0.025 = 1 : 1, so the formula is MgO.

Worked example — a compound of nitrogen and hydrogen contains 1.4 g of nitrogen and 0.3 g of hydrogen. Find its formula.

• Moles N = 1.4 ÷ 14 = 0.1
• Moles H = 0.3 ÷ 1 = 0.3

The ratio N : H is 0.1 : 0.3 = 1 : 3, so the formula is NH₃. This is exactly how empirical formulae and balancing numbers are found from experimental data.

Reactants are rarely mixed in the exact ratio the equation demands. When one runs out first, the reaction stops, and no more product can form however much of the other reactant remains.

The limiting reactant is the one used up completely. It controls the amount of product. The other reactant, present in more than enough, is the reactant in excess.

To decide which reactant is limiting, convert both masses to moles and compare them against the mole ratio in the equation. The one that provides fewer moles than the ratio requires is limiting.

Chemists often add one reactant in excess on purpose, to make sure the more valuable or slower-reacting substance is used up fully. The amount of product then depends only on the limiting reactant, so that is always the basis for the calculation.

Worked example — 28 g of nitrogen reacts with 3 g of hydrogen. What is the maximum mass of ammonia formed?

${\text{N}}_2+3{\text{H}}_2\to 2{\text{NH}}_3$

Step 1 — moles of each reactant:

• Moles N₂ (Mr = 28) = 28 ÷ 28 = 1 mol
• Moles H₂ (Mr = 2) = 3 ÷ 2 = 1.5 mol

Step 2 — identify the limiting reactant. The equation needs 3 mol of H₂ for every 1 mol of N₂. To react with 1 mol of N₂ you would need 3 mol of H₂, but only 1.5 mol is present. Hydrogen runs out first, so hydrogen is limiting and nitrogen is in excess.

Step 3 — base the product on the limiting reactant. H₂ : NH₃ is 3 : 2, so moles of NH₃ = 1.5 × (2 ÷ 3) = 1.0 mol.

Step 4 — mass of NH₃ (Mr = 14 + 3 = 17):

${\text{mass NH}}_3=17\times 1.0=17\text{g}$

The maximum mass of ammonia is 17 g. Using the nitrogen instead would give a wrong, larger answer, because there is not enough hydrogen to react with all of it.

1. Using masses directly in the mole ratio

The multipliers give a ratio of moles, not masses. Convert every mass to moles before applying the ratio, then convert the answer back to a mass.

2. Applying the ratio the wrong way round

Check whether the substance you want has more or fewer moles than the one you know. For Al : Al₂O₃ = 2 : 1 you halve the moles; getting the direction wrong doubles or halves the answer.

3. Basing the product on the reactant in excess

The amount of product is set by the limiting reactant. Always identify which reactant runs out first and calculate from that one.

4. Forgetting to compare against the equation's ratio

Having more moles of a reactant does not make it in excess by itself. Compare the moles you have with the ratio the equation demands before deciding which is limiting.

5. Not simplifying the mole ratio when balancing

When balancing from masses, write the moles as the simplest whole-number ratio. A ratio of 0.1 : 0.3 must be simplified to 1 : 3 to give the formula NH₃.