Lesson 4 · 30 min

Limiting reagents and yield

Adjust the reactants in the reaction mixer, see which runs out first and how much is left, and work out the theoretical and percent yield.

In this lesson

Explain what the limiting and excess reagents are
Find the limiting reagent with nᵢ/νᵢ and the amounts of everything at the end
Calculate the theoretical yield, the excess left over and the percent yield
Organise the working in a before–change–after table

Making sandwiches: each needs 2 slices of bread and 1 slice of cheese. With 10 slices of bread and 4 of cheese, how many can you make? The bread is enough for 5, the cheese for only 4 — so you make 4 and have 2 slices of bread left. The cheese sets the output: it is the limiting ingredient. Reactions are exactly the same. Reactants are rarely mixed in exactly the ratio of the equation; the one that runs out first — the limiting reagent — decides how much product forms, and the others are left over in excess.

Predict first

5 mol of $\ce{H2}$ and 2 mol of $\ce{O2}$ are mixed and ignited ( $\ce{2H2} + \ce{O2} \mathrel{\ce{->}} \ce{2H2O}$ ). What is left in the vessel at the end?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Only 5 mol of water 4 mol of water and 1 mol of

\ce{H2}

4 mol of water and 0.5 mol of

\ce{O2}

2 mol of water and 3 mol of

\ce{H2}

Finding the limiting reagent

Comparing masses will not do, and neither will comparing moles — divide each reactant’s moles by its coefficient. The ratio $n_i/\nu_i$ says “how many rounds of the reaction this reactant can support”; the smallest one is the limiting reagent, and its value is the maximum extent of reaction, $\xi$ .

Once you know $\xi$ , everything follows: each reactant loses $\nu_i\xi$ , each product gains $\nu_j\xi$ , and the excess reagent has $n_i - \nu_i\xi$ left. A three-row before–change–after table keeps it tidy.

\xi = \min_i \frac{n_i}{\nu_i} \qquad m_\text{theor} = \nu_\mathrm{P}\,\xi\,M_\mathrm{P} \qquad \text{percent yield} = \frac{m_\text{actual}}{m_\text{theor}} \times 100\%

Extent of reaction, theoretical yield and percent yield.

$\xi$: extent of reaction (mol): how many “rounds” of the equation occurred
$n_i,\ \nu_i$: initial moles of reactant i and its coefficient
$m_\text{theor}$: theoretical yield: mass of product P if the limiting reagent reacts completely
$m_\text{actual}$: actual yield: the mass really obtained

Limiting reagent · excess · theoretical and percent yield

Reaction mixer: which runs out first?

Reaction

\ce{N2(g) + 3H2(g) -> 2NH3(g)}

Amount of N₂ 2 mol · 56.0 g Amount of H₂ 5 mol · 10.1 g

A tiny sample in the same proportions: 3 particles drawn per mol N₂ × 1H₂ × 0 NH₃ × 10

Progress (reaction events so far) 5 / 5

Limiting reagent: H₂ N₂ in excess · 0.333 mol

Before · Change · After (mol) · ξ = 1.67 mol
Species	Before	Change	After	Mass after / g
$\ce{N2}$	2	−1.67	0.333	9.34
$\ce{H2}$	5	−5.00	0.00	0.00
$\ce{NH3}$	0	+3.33	3.33	56.77

Theoretical and percent yield

Theoretical yield (NH₃) 56.77 g

Actual yield 48.25 g

Percent yield 85.0 %

In ammonia synthesis

\ce{H2}

has a coefficient of 3, so it can be limiting even when there is more of it. Move the sliders to find the exactly stoichiometric mixture, then switch to zinc and hydrochloric acid and see the yield slider warn you above 100%.

Worked example

Making ammonia: which runs out?

28.0 g of $\ce{N2}$ reacts with 8.0 g of $\ce{H2}$ to form $\ce{NH3}$ ( $\ce{N2} + \ce{3H2} \mathrel{\ce{->}} \ce{2NH3}$ ). Find the limiting reagent, the theoretical yield and the excess left over. If 25.5 g of $\ce{NH3}$ is actually obtained, what is the percent yield?

Concept check

0.10 mol of zinc reacts with 0.15 mol of HCl ( $\ce{Zn} + \ce{2HCl} \mathrel{\ce{->}} \ce{ZnCl2} + \ce{H2}$ ). Which is limiting?

Zinc HCl Neither — exactly stoichiometric Cannot tell

Try the calculation

10.0 g of $\ce{H2}$ reacts with 64.0 g of $\ce{O2}$ . What is the maximum mass of water?

Key takeaways

The limiting reagent has the smallest $n_i/\nu_i$ ; it sets the extent $\xi$ and the amount of product.
The excess reagent keeps $n_i - \nu_i\xi$ ; a before–change–after table organises the work.
The theoretical yield assumes the limiting reagent reacts completely; percent yield = actual ÷ theoretical × 100%.
Yields below 100% come from equilibrium, side reactions and handling losses; above 100% usually means an impure product.

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Data and references

IUPAC Gold Book, “extent of reaction, ξ”.
N. N. Greenwood, A. Earnshaw, Chemistry of the Elements, 2nd ed., Ch. 11 (the Haber–Bosch process).
S. S. Zumdahl, S. A. Zumdahl, Chemistry, 10th ed., Section 3.11 (limiting reactants).