Limiting reagents and percentage yield

Mass-to-mass stoichiometry using the mole ratio, identifying the limiting reagent and the amount of excess reagent left over, theoretical, actual and percentage yield and why yields fall short, and an investigation of gas volume against mass of magnesium.

A café has 40 slices of bread and 12 slices of cheese. Each toasted sandwich needs 2 slices of bread and 1 slice of cheese. However much bread is in the cupboard, the cheese runs out after 12 sandwiches. Chemical reactions work exactly the same way, except that the "slices" are moles and the recipe is the balanced equation.

Mass-to-mass calculations

Every stoichiometry problem follows the same three steps:

Known to moles: n = m ÷ M for the substance you know about.
Mole ratio: use the coefficients to find moles of the substance you want. $n(\text{want}) = n(\text{know}) \times \dfrac{\text{coefficient of want}}{\text{coefficient of know}}$
Moles to answer: m = nM (or a volume or number of particles).

Iron from iron ore

In a blast furnace, Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g). What mass of iron can be made from 1.00 tonne of Fe₂O₃?

1.00 t = 1.00 × 10⁶ g. M(Fe₂O₃) = 2 × 55.85 + 3 × 16.00 = 159.70 g mol⁻¹, so n = 1.00 × 10⁶ ÷ 159.70 = 6262 mol.

Ratio Fe : Fe₂O₃ = 2 : 1, so n(Fe) = 2 × 6262 = 12 524 mol.

m(Fe) = 12 524 × 55.85 = 6.99 × 10⁵ g = 0.699 t. Western Australia's Pilbara exports hundreds of millions of tonnes of iron ore a year, and calculations like this decide how much steel each shipment can make.

Limiting and excess reagents

When amounts of two reactants are given, one usually runs out first. That one is the limiting reagent: it limits the amount of product. The other is in excess, and some of it is left over.

Five H₂ molecules and two O₂ molecules. The two O₂ can react with only four H₂, so O₂ is limiting and one H₂ is left over.

To find the limiting reagent reliably, divide the moles of each reactant by its coefficient. The smallest value marks the limiting reagent, and that value tells you how many "lots" of the equation can happen. Comparing raw masses, or even raw moles, will mislead you whenever the coefficients are different.

Aluminium and chlorine

10.0 g of aluminium reacts with 30.0 g of chlorine: 2Al(s) + 3Cl₂(g) → 2AlCl₃(s). Find the limiting reagent, the mass of AlCl₃ formed and the mass of excess reagent left over.

n(Al) = 10.0 ÷ 26.98 = 0.3706 mol; n(Cl₂) = 30.0 ÷ 70.90 = 0.4231 mol.

Divide by coefficients: Al 0.3706 ÷ 2 = 0.1853; Cl₂ 0.4231 ÷ 3 = 0.1410. Cl₂ is smaller, so chlorine is limiting.

n(AlCl₃) = 0.4231 × $\tfrac{2}{3}$ = 0.2821 mol; m = 0.2821 × 133.33 = 37.6 g.

Al used = 0.4231 × $\tfrac{2}{3}$ = 0.2821 mol = 7.61 g, so 10.0 − 7.61 = 2.39 g of Al is left over. Check: 37.6 + 2.39 = 40.0 g = 10.0 + 30.0 g ✓ (mass is conserved).

Percentage yield

The theoretical yield is the mass of product calculated from the limiting reagent, assuming it all reacts and nothing is lost.
The actual yield is the mass you really collect.

$$\%\ \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$$

Real yields are below 100% because of incomplete reactions (including reversible reactions that reach equilibrium, Year 12 Module 5), side reactions that make other products, product left on glassware or filter paper, and losses during purification. A yield above 100% means the product is impure or still wet.

Where the missing product goes in a typical student synthesis.

Burning magnesium

5.00 g of magnesium is burned and 7.80 g of magnesium oxide is collected. 2Mg(s) + O₂(g) → 2MgO(s). Calculate the percentage yield.

n(Mg) = 5.00 ÷ 24.31 = 0.2057 mol; ratio 1 : 1, so n(MgO) = 0.2057 mol (oxygen from the air is in excess).

Theoretical yield = 0.2057 × 40.31 = 8.29 g.

% yield = 7.80 ÷ 8.29 × 100 = 94.1%.

Working backwards from a known yield

Aspirin is made from salicylic acid (M = 138.1) in a 1 : 1 ratio, with a typical yield of 85%. What mass of salicylic acid is needed to make 50.0 g of aspirin (M = 180.2)?

Theoretical yield needed = 50.0 ÷ 0.85 = 58.8 g of aspirin = 58.8 ÷ 180.2 = 0.326 mol.

n(salicylic acid) = 0.326 mol, so m = 0.326 × 138.1 = 45.1 g.

InteractiveLimiting-reagent mixer

Try this

With 4.00 g of H₂ and 16.00 g of O₂, press React. Which is limiting, even though there is four times the mass of O₂? Check: 18.02 g of water and 1.98 g of H₂ left over.
Find the mass of O₂ that reacts exactly with 4.00 g of H₂, so that nothing is left over.
Choose N₂ + 3H₂ ⇌ 2NH₃ with 28.00 g of N₂ and 9.00 g of H₂. The theoretical yield is 34.04 g. Turn on Real lab: why is the yield only 15%?
Choose Fe + S → FeS and set the masses so both are used up exactly (hint: equal moles).

Investigation: how much hydrogen? Finding the limiting reagent

Aim: to investigate how the volume of hydrogen produced depends on the mass of magnesium when the amount of acid is fixed.

Materials: six 250 mL conical flasks, six balloons, 0.200 mol L⁻¹ HCl, 50 mL measuring cylinder, magnesium ribbon, balance (±0.001 g), string and ruler (or a gas syringe).

Method: put 50.0 mL of acid in each flask. Weigh magnesium samples from 0.020 g to 0.220 g. Place each sample in a balloon, stretch the balloon over the flask and tip the magnesium in. When fizzing stops, measure the volume of gas (with a gas syringe, or estimate from the balloon circumference).

Results and analysis: the volume rises in proportion to the mass of magnesium, then levels off.

Volume of hydrogen against mass of magnesium for 50.0 mL of 0.200 mol L⁻¹ HCl. The break in the graph is where the limiting reagent changes.

n(HCl) = 0.200 × 0.0500 = 0.0100 mol, which reacts with 0.00500 mol of Mg (0.122 g) and makes at most 0.00500 mol of H₂ (124 mL at 25 °C and 100 kPa, see 2I). Below 0.122 g magnesium is limiting; above it the acid is limiting and extra magnesium just sits there unreacted.

Validity: keep the volume and concentration of acid and the temperature the same. Balloons are only rough measures of volume; a gas syringe is more accurate. Magnesium ribbon with a white oxide coating should be cleaned with steel wool so its mass is all magnesium.

Safety: hydrogen is flammable: no flames nearby. Wear safety glasses.

Test yourself

N₂(g) + 3H₂(g) → 2NH₃(g). 2.0 mol of N₂ is mixed with 4.5 mol of H₂. Which is limiting and how much NH₃ can form?
What mass of CO₂ is produced when 8.0 g of methane burns completely? CH₄ + 2O₂ → CO₂ + 2H₂O
A student's yield of a dried product is 104%. The most likely reason is that
6.0 g of magnesium is added to 0.40 mol of HCl. Mg + 2HCl → MgCl₂ + H₂. What amount of the excess reagent remains?
On a graph of volume of H₂ against mass of Mg added to a fixed amount of acid, what does the horizontal section show?

Put the steps for a limiting-reagent problem in order.

Sort the effect on the measured percentage yield.

The reactant that is completely used up is the reagent. It has the value of moles divided by coefficient. The mass of product calculated from it is the yield. Percentage yield is actual yield divided by theoretical yield, multiplied by .

6 marks

Titanium is produced by the Kroll process: TiCl₄(g) + 2Mg(l) → Ti(s) + 2MgCl₂(l). In one batch 3.54 × 10⁴ g of TiCl₄ reacts with 1.13 × 10⁴ g of magnesium, and 7.91 × 10³ g of titanium is obtained. (a) Identify the limiting reagent. (b) Calculate the percentage yield. (c) Suggest one reason the yield is below 100%.

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