Limiting and Excess Reactants

Sandwich math

One sandwich takes 2 slices of bread and 1 slice of cheese. With 10 slices of bread and 3 slices of cheese, how many sandwiches can you make? The bread is enough for 5, but the cheese is enough for only 3. So you make 3 sandwiches and have 4 slices of bread left over. The cheese limits the number of sandwiches, and the bread is in excess. Reactions work the same way: reactants are rarely mixed in exactly the ratio of the equation, so one runs out first.

The reactant that is used up first, and so sets the maximum amount of product, is the limiting reactant限量试剂 (also called the limiting reagent). Any reactant left over when the reaction stops is an excess reactant过量试剂.

The amount of product must be calculated from the limiting reactant. Starting from an excess reactant gives an amount of product you can never get.
Chinese problems rarely use the term “limiting reactant.” They say a reactant is 过量 (in excess), 足量 (in sufficient amount), or 不足 (insufficient) and ask you to calculate from the insufficient one. When the last lesson said “excess hydrochloric acid,” it was telling you that zinc is the limiting reactant.

Finding the limiting reactant

When the amounts of two reactants are given, convert both to moles, then use either method:

Compare mol ÷ coefficient. Divide each reactant’s moles by its coefficient in the equation. The smaller result runs out first, so it is the limiting reactant. Chinese textbooks usually use this method.
Compare the product each could make. Assume each reactant reacts completely and calculate how much product it would make. The reactant that makes less is limiting, and that smaller amount is the maximum amount of product. US textbooks usually use this method.

Both methods always agree. Once you know the limiting reactant:

Product formed = the amount calculated from the limiting reactant.
Excess reactant used = the amount calculated from the limiting reactant and the mole ratio; excess left over = starting amount − amount used.

This lesson uses the atomic masses on the AP periodic table: H 1.008, C 12.01, O 16.00, Na 22.99, Cl 35.45, Fe 55.85, and Zn 65.38.

Start with particles. Change the number of each kind of molecule, predict which runs out first, and then press React:

Try it: particle view — which runs out first?

Before

After

HOC

Now work with masses. Change the masses of the two reactants and watch which one is limiting and how much of the other is left:

Try it: find the limiting reactant from masses

Product:

Reactant	Mass (g)	M (g/mol)	mol	mol ÷ coefficient	Product it could make

Two ways to decide: compare mol ÷ coefficient for each reactant (the smaller one runs out first, the usual method in Chinese textbooks), or calculate how much product each reactant could make (the one that makes less is limiting, the usual method in US textbooks). Both give the same answer.

Example 1: compare mol ÷ coefficient

A mixture of 2.40 g H₂ and 16.0 g O₂ is ignited and reacts completely. Which reactant is limiting? What mass of water forms, and how much of which gas is left over?

\ce{2H2(g) + O2(g) -> 2H2O(l)}

Step-by-step solution

Convert to moles.
$\begin{aligned} n(\ce{H2}) &= 2.40 / 2.016 \\ &= 1.190\ \text{mol} \\ n(\ce{O2}) &= 16.0 / 32.00 \\ &= 0.500\ \text{mol} \end{aligned}$
Compare moles ÷ coefficient. The smaller result runs out first:
$\begin{aligned} \ce{H2}: \ & 1.190 / 2 = 0.595 \\ \ce{O2}: \ & 0.500 / 1 = 0.500 \end{aligned}$
O₂ is the limiting reactant.
Water from O₂.
$\begin{aligned} n(\ce{H2O}) &= 0.500 \times 2 \\ &= 1.000\ \text{mol} \\ m(\ce{H2O}) &= 1.000 \times 18.02 \\ &= \boxed{18.0\ \text{g}} \end{aligned}$
H₂ used and left over. H₂ used = 0.500 × 2 = 1.000 mol, which is 2.016 g:
$2.40 - 2.016 = \boxed{0.38\ \text{g}}\ \ce{H2}$
(The subtraction keeps two decimal places.)
Check mass conservation. 2.40 g + 16.0 g ≈ 18.0 g + 0.38 g.
A common mistake. Comparing moles directly (0.500 < 1.190, so O₂ is limiting) and getting the right answer by luck. Divide each by its coefficient before comparing.

Example 2: compare the product each could make

6.50 g of zinc reacts with hydrochloric acid containing 3.65 g of HCl. What mass of hydrogen forms? Which reactant is left over, and how much?

\ce{Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)}

Step-by-step solution

Convert to moles.
$\begin{aligned} n(\ce{Zn}) &= 6.50 / 65.38 \\ &= 0.09942\ \text{mol} \\ n(\ce{HCl}) &= 3.65 / 36.46 \\ &= 0.1001\ \text{mol} \end{aligned}$
How much H₂ could each make, if it all reacted?
$\begin{aligned} \text{from Zn: } & 0.09942 \times \tfrac{1}{1} = 0.09942 \\ \text{from HCl: } & 0.1001 \times \tfrac{1}{2} = 0.05005 \end{aligned}$
(in mol of H₂)
Take the smaller amount. HCl makes less, so HCl is limiting:
$\begin{aligned} m(\ce{H2}) &= 0.05005 \times 2.016 \\ &= \boxed{0.101\ \text{g}} \end{aligned}$
Zinc left over. Zn used = 0.05005 × 65.38 = 3.272 g:
$6.50 - 3.272 = \boxed{3.23\ \text{g}}\ \ce{Zn}$
Think about it. The two reactants started with nearly equal moles, but HCl runs out first because each Zn atom needs two HCl.

Practice: for 2H₂(g) + O₂(g) → 2H₂O(l), which reactant is limiting?

4 mol H₂ and 1 mol O₂
4 mol H₂ and 3 mol O₂
6 mol H₂ and 3 mol O₂
1.5 mol H₂ and 1 mol O₂
10 molecules of H₂ and 4 molecules of O₂
4.032 g H₂ and 32.00 g O₂
2.0 g H₂ and 32 g O₂
8.0 g H₂ and 32 g O₂

Common mistakes

Mistake 1: “The reactant with the smaller mass is limiting.” In Example 1, there was only 2.40 g of H₂, far less than the 16.0 g of O₂, yet H₂ was in excess. Compare moles, not grams.

Mistake 2: “The reactant with fewer moles is limiting.” The coefficients matter too. With 1.5 mol H₂ and 1 mol O₂, O₂ has fewer moles, but 1 mol O₂ needs 2 mol H₂, so H₂ is the one that runs short. Compare mol ÷ coefficient.

Mistake 3: “Either product amount is fine.” Only the smaller amount, from the limiting reactant, can actually form. The larger amount assumes the excess reactant reacts completely, which it never does.

Quick check

2CO(g) + O₂(g) → 2CO₂(g). 3 mol CO and 2 mol O₂ react completely. Which is limiting, and how much CO₂ forms?

2Na(s) + Cl₂(g) → 2NaCl(s). 4.60 g Na reacts with 10.0 g Cl₂. What mass of NaCl forms?

Summary

The reactant that runs out first is the limiting reactant and sets the maximum amount of product. Any reactant left over is in excess.
To find it, compare mol ÷ coefficient (the smaller runs out first), or compare the product each reactant could make (the smaller amount wins). Never compare grams or raw moles.
Calculate the product from the limiting reactant. Excess left over = starting amount − amount used, where the amount used comes from the limiting reactant.

On to AP (for later)

Limiting reactants are a key topic in AP Chemistry Unit 4. AP questions often use particle diagrams: given a container of molecules before the reaction, draw the container afterward, or decide which reactant is in excess. Free-response questions may ask you to justify your answer, so state the comparison, for example: “O₂ can make only 1.000 mol H₂O, less than the 1.190 mol H₂ can make, so O₂ is limiting.”

Limiting and Excess Reactants限量试剂与过量试剂

Learning goals

Before you start

Sandwich math