Use mole ratios from balanced equations to calculate reacting masses and gas volumes, identify the limiting reagent, and calculate percentage yield

What this dot point is asking

SEAB wants you to use the mole ratios from a balanced equation to calculate reacting masses and the volumes of gases produced, to identify the limiting reagent when two amounts are given, and to calculate percentage yield. These calculations turn a balanced equation into a numerical answer, and they appear in every quantitative question on the written paper.

The answer

The three-step reacting-mass method

Almost every reacting-mass question follows the same three steps:

1. Moles of what you know. Convert the given mass to moles using $n = m/M_r$.
2. Mole ratio. Use the coefficients in the balanced equation to find the moles of what you want.
3. Answer. Convert those moles to the quantity asked for (mass with $m = n \times M_r$, or gas volume, see below).

Get the balanced equation right first, because the whole calculation rests on the mole ratio it provides.

Gas volumes at room temperature and pressure

At room temperature and pressure (r.t.p.), one mole of any gas occupies $24.0\ \text{dm}^3$ (or $24\,000\ \text{cm}^3$). So:

This lets you find the volume of a gas produced: find its moles by the three-step method, then multiply by $24.0\ \text{dm}^3$. The molar volume is the same for every gas because gas particles are far apart, so the size of the particles barely matters.

The limiting reagent

When the amounts of two reactants are both given, one usually runs out first; this is the limiting reagent, and it controls how much product forms. The other reactant is in excess. To find which is limiting:

1. Find the moles of each reactant.
2. Divide each by its coefficient in the equation.
3. The smallest result is the limiting reagent.

Always base the product calculation on the limiting reagent, never on the one in excess.

Percentage yield

The theoretical yield is the mass (or moles) of product the equation predicts. The actual yield is what is really obtained, which is usually less, because reactions may not finish, products are lost in handling, or side reactions occur. The percentage yield compares the two:

Examples in context

Example 1. Working out a fertiliser yield. A manufacturer calculates the theoretical mass of ammonium sulfate from the reacting masses, then compares the mass actually produced to find the percentage yield. A yield well below $100\%$ flags losses in the process, so the calculation guides where to improve efficiency.

Example 2. Predicting gas for a reaction. Before running a reaction that gives off hydrogen, a chemist calculates the expected volume at r.t.p. using the molar gas volume, so the right size of gas syringe or collection vessel is chosen. The reacting-mass method links the solid reactant to the gas volume produced.

Try this

Q1. State the three steps of a reacting-mass calculation. [1 mark]

• Cue. Moles of the known substance, mole ratio from the balanced equation, then convert to the quantity asked for.

Q2. $0.20$ mol of a gas is collected at r.t.p. Calculate its volume. (Molar gas volume $= 24.0\ \text{dm}^3$.) [1 mark]

• Cue. $V = 0.20 \times 24.0 = 4.8\ \text{dm}^3$.

Q3. $4.8\ \text{g}$ of magnesium ($A_r = 24$) and $4.8\ \text{g}$ of oxygen ($\text{O}_2$, $M_r = 32$) react by $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. Identify the limiting reagent. [3 marks]

• Cue. $n(\text{Mg}) = 0.20$, divide by $2$ gives $0.10$; $n(\text{O}_2) = 0.15$, divide by $1$ gives $0.15$; magnesium has the smaller value, so magnesium is limiting.