Limiting Reagent Reactions: HSC Chemistry Explained

Suppose you mix two reactants in exactly the masses written on the bottle, heat them, and wait until the reaction stops. One reactant is completely gone. The other is still sitting there. Which one decided how much product you could make?

A tempting answer is “the reactant with the smaller mass”. That sounds reasonable, but it can be completely wrong. Chemical equations compare particles, and we count particles using moles, not grams.

Before reading on, predict this: if a reaction needs 2 mol of A for every 1 mol of B, and you mix 3 mol of A with 2 mol of B, which reactant runs out first?

You have enough B to react with 4 mol of A, but only 3 mol of A are available. So A runs out first. A is the limiting reagent. B is left over in excess.

That one idea controls three common HSC calculations:

• identifying the limiting reagent
• calculating the theoretical amount of product
• calculating how much excess reactant remains

01The basic picture: a chemical reaction is a ratio problem

Imagine you’re making very simple sandwiches. Each sandwich needs 2 slices of bread and 1 slice of cheese.

If you have:

• 10 slices of bread
• 8 slices of cheese

you can’t make 8 sandwiches. The cheese would allow 8, but the bread only allows 5.

The bread limits production.

A balanced chemical equation works in the same way. Its coefficients tell you the required mole ratio.

For example:

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

This means:

• 2 mol of \(\ce{H2}\) react with 1 mol of \(\ce{O2}\)
• those amounts produce 2 mol of \(\ce{H2O}\)

The analogy is useful because both situations involve fixed ratios. It breaks down because molecules aren’t actually lining up like sandwich ingredients. Chemical reactions involve collisions, bond breaking, and bond formation. The ratio still comes from the balanced equation.

02What is the limiting reagent?

The limiting reagent is the reactant that is completely consumed first according to the stoichiometric ratio.

Once it has been used up, no more product can form, even if another reactant remains.

The other reactant is an excess reagent. Some of it reacts, but there is more present than the reaction requires.

Notice what the definition does not say. The limiting reagent is not necessarily:

• the reactant with the smaller mass
• the reactant with the smaller number of moles
• the reactant with the smaller coefficient

You must compare the amount available with the amount required by the balanced equation.

03The reliable method

For most HSC limiting reagent problems, use this sequence:

1. Balance the chemical equation.
2. Convert each reactant amount to moles.
3. Compare the available moles using the mole ratio.
4. Identify the limiting reagent.
5. Use the limiting reagent to calculate product.
6. If required, calculate how much excess reagent reacted.
7. Subtract that from the initial amount of excess reagent.

If you need a refresher on converting mass to moles, see Moles from Mass and Molar Mass: HSC Chemistry Guide.

The key conversion is:

\(n = \frac{m}{M}\)

where:

• \(n\) is amount in moles (mol)
• \(m\) is mass in grams (g)
• \(M\) is molar mass in grams per mole (g mol\(^{-1}\))

04How do you actually compare the reactants?

Consider:

\(\ce{2A + 3B -> products}\)

Suppose you have 4 mol of A and 9 mol of B.

Looking only at moles, A is smaller. But that isn’t yet enough.

The reaction requires:

\(\frac{n(A)}{n(B)} = \frac{2}{3}\)

If all 4 mol of A react, the amount of B required is:

\(n(B) = 4 \times \frac{3}{2} = 6\text{ mol}\)

You have 9 mol of B, so there is more than enough B.

Therefore A is limiting.

A faster comparison: divide by the coefficient

There is a useful shortcut. Divide the number of moles of each reactant by its coefficient in the balanced equation.

For:

\(\ce{2A + 3B -> products}\)

compare:

\(\frac{n(A)}{2}\)

with:

\(\frac{n(B)}{3}\)

The smaller value corresponds to the limiting reagent.

For 4 mol A and 9 mol B:

\(\frac{4}{2}=2\)

and

\(\frac{9}{3}=3\)

A has the smaller value, so A is limiting.

Why does this work? Each result tells you how many complete “reaction batches” that reactant could support.

This shortcut is excellent once the equation is balanced. If the equation is not balanced, the comparison is meaningless.

06Why the limiting reagent must control the product calculation

Once you’ve identified the limiting reagent, use it to calculate product.

Don’t calculate product from the excess reagent unless you deliberately account for the fact that some of it cannot react.

Suppose the limiting reagent can form 0.40 mol of product, while the excess reagent would apparently allow 0.65 mol.

Could 0.65 mol actually form?

No. The reaction stops once the limiting reagent is gone. The larger prediction assumes that more limiting reagent exists than is actually present.

The theoretical yield is therefore the maximum amount of product predicted from the limiting reagent, assuming complete reaction and no product loss.

It isn’t automatically the amount you would collect in a real experiment. Actual yield can be smaller because of incomplete reactions, competing reactions, transfer losses, purification losses, and other experimental limitations.

08The misconception that causes most mistakes

A student sees:

• 5 g of one reactant
• 20 g of another

and immediately says the 5 g reactant must be limiting.

Why does that feel plausible? Because in ordinary life, “less stuff” often runs out first.

Chemically, though, grams aren’t a particle count. A mole of one substance can have a completely different mass from a mole of another.

Even comparing moles directly can fail.

Consider:

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

Suppose you have:

• 1.0 mol of \(\ce{N2}\)
• 2.0 mol of \(\ce{H2}\)

Hydrogen has more moles, but it is still limiting.

One mole of nitrogen requires 3 mol of hydrogen. There are only 2 mol available.

The correct comparison is therefore not “which number is smaller?” It is “which reactant runs out first when the balanced ratio is enforced?”

09Two valid ways to identify the limiting reagent

You don’t have to use exactly one method every time.

For example, with:

\(\ce{2A + 5B -> products}\)

and 0.60 mol A plus 1.20 mol B:

Using \(n/\text{coefficient}\):

\(\frac{0.60}{2}=0.30\)

\(\frac{1.20}{5}=0.24\)

B is limiting.

Using the required amount method, 0.60 mol of A would need:

\(0.60\times\frac{5}{2}=1.50\text{ mol B}\)

but only 1.20 mol B is available. Same conclusion.

If two methods disagree, the chemistry hasn’t become mysterious. You’ve made an arithmetic, conversion, or balancing error somewhere.

10How to calculate excess reagent remaining

Once the limiting reagent has been identified, this is a separate stoichiometry calculation.

The logic is:

\(\text{excess remaining} = \text{excess initially present} – \text{excess that reacted}\)

The important part is finding how much excess reagent actually reacted.

Use the limiting reagent and the balanced equation to calculate that amount.

Do not simply subtract the amount of limiting reagent from the amount of excess reagent. Their mole ratio might not be 1:1, and subtracting masses of different substances has no chemical meaning.

A good written solution usually follows this chain:

\(\text{limiting reagent} \rightarrow \text{moles of excess consumed} \rightarrow \text{moles of excess remaining} \rightarrow \text{mass remaining, if required}\)

11What if the reactants are present in exactly the required ratio?

Suppose:

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

and you have exactly:

• 2.00 mol \(\ce{H2}\)
• 1.00 mol \(\ce{O2}\)

Which reactant is in excess?

Neither.

They are present in the exact stoichiometric ratio. Under the ideal model, both are completely consumed.

Using the shortcut confirms this:

\(\frac{2.00}{2}=1.00\)

and:

\(\frac{1.00}{1}=1.00\)

The values are equal.

So don’t force every problem into “one limiting, one excess”. Sometimes neither reactant is in excess.

12Questions and solutions

Question 1

Magnesium reacts with hydrochloric acid according to:

\(\ce{Mg + 2HCl -> MgCl2 + H2}\)

A mixture contains 0.150 mol of \(\ce{Mg}\) and 0.240 mol of \(\ce{HCl}\).

Identify the limiting reagent and calculate the amount, in moles, of \(\ce{H2}\) that can theoretically form.

Solution 1

\(\ce{HCl}\) is the limiting reagent, and 0.120 mol of \(\ce{H2}\) can theoretically form.

Compare the reactant amounts after dividing by their coefficients:

\(\begin{aligned} \frac{n(\ce{Mg})}{1}&=\frac{0.150}{1}=0.150 \\ \frac{n(\ce{HCl})}{2}&=\frac{0.240}{2}=0.120 \end{aligned}\)

The smaller value belongs to \(\ce{HCl}\), so hydrochloric acid is limiting.

From:

\(\ce{Mg + 2HCl -> MgCl2 + H2}\)

2 mol of \(\ce{HCl}\) produce 1 mol of \(\ce{H2}\).

Therefore:

\(\begin{aligned} n(\ce{H2}) &=0.240\times\frac{1}{2} =0.120\text{ mol} \end{aligned}\)

The reaction stops once the 0.240 mol of \(\ce{HCl}\) has been consumed, even though some magnesium remains.

Question 2

Calcium reacts with water according to:

\(\ce{Ca + 2H2O -> Ca(OH)2 + H2}\)

A sample contains 4.01 g of calcium and 5.40 g of water.

Use:

• \(M(\ce{Ca})=40.08\text{ g mol}^{-1}\)
• \(M(\ce{H2O})=18.02\text{ g mol}^{-1}\)

Identify the limiting reagent and calculate the amount of excess reactant remaining in moles.

Solution 2

Calcium is the limiting reagent, and 0.100 mol of water remains after the reaction.

First convert both masses to moles:

\(\begin{aligned} n(\ce{Ca}) &=\frac{4.01}{40.08} =0.100\text{ mol} \\ n(\ce{H2O}) &=\frac{5.40}{18.02} =0.300\text{ mol} \end{aligned}\)

Now compare each value with its coefficient:

\(\begin{aligned} \frac{0.100}{1}&=0.100 \\ \frac{0.300}{2}&=0.150 \end{aligned}\)

Calcium gives the smaller value, so it is limiting.

The equation requires 2 mol of water for every 1 mol of calcium. Therefore the amount of water consumed is:

\(\begin{aligned} n(\ce{H2O\ reacted}) &=0.100\times\frac{2}{1} =0.200\text{ mol} \end{aligned}\)

The amount remaining is:

\(\begin{aligned} n(\ce{H2O\ remaining}) &=0.300-0.200 =0.100\text{ mol} \end{aligned}\)

The trap here is to assume that water must be limiting because each mole of calcium needs two moles of water. There are actually enough water molecules present for all the calcium to react.

Question 3

Iron reacts with oxygen according to:

\(\ce{4Fe + 3O2 -> 2Fe2O3}\)

A reaction vessel initially contains 11.2 g of iron and 6.40 g of oxygen gas.

Use:

• \(M(\ce{Fe})=55.85\text{ g mol}^{-1}\)
• \(M(\ce{O2})=32.00\text{ g mol}^{-1}\)
• \(M(\ce{Fe2O3})=159.7\text{ g mol}^{-1}\)

Determine the limiting reagent and theoretical mass of \(\ce{Fe2O3}\).

Solution 3

Iron is the limiting reagent, and the theoretical mass of \(\ce{Fe2O3}\) is 16.0 g.

Convert each reactant to moles:

\(\begin{aligned} n(\ce{Fe}) &=\frac{11.2}{55.85} =0.2005\text{ mol} \\ n(\ce{O2}) &=\frac{6.40}{32.00} =0.2000\text{ mol} \end{aligned}\)

The mole amounts are almost identical, but the balanced equation does not use iron and oxygen in a 1:1 ratio.

Compare moles divided by coefficients:

\(\begin{aligned} \frac{0.2005}{4}&=0.0501 \\ \frac{0.2000}{3}&=0.0667 \end{aligned}\)

Iron has the smaller value, so iron is limiting.

From the balanced equation:

\(4\text{ mol }\ce{Fe}:2\text{ mol }\ce{Fe2O3}\)

Therefore:

\(\begin{aligned} n(\ce{Fe2O3}) &=0.2005\times\frac{2}{4} =0.1003\text{ mol} \end{aligned}\)

Convert to mass:

\(\begin{aligned} m(\ce{Fe2O3}) &=0.1003\times159.7 =16.0\text{ g} \end{aligned}\)

The important trap is the nearly equal number of moles. Equal mole amounts do not imply a stoichiometric mixture when the coefficients are different.

Question 4

Phosphorus reacts with chlorine according to:

\(\ce{P4 + 6Cl2 -> 4PCl3}\)

A sealed vessel contains 0.0800 mol of \(\ce{P4}\) and 0.480 mol of \(\ce{Cl2}\).

A student claims that chlorine must be the limiting reagent because six times as many moles of chlorine are required.

Determine whether the claim is correct, identify any excess reactant, and calculate the theoretical amount of \(\ce{PCl3}\).

Solution 4

The student’s claim is incorrect. Neither reactant is in excess, and 0.320 mol of \(\ce{PCl3}\) can theoretically form.

The equation requires:

\(1\text{ mol }\ce{P4}:6\text{ mol }\ce{Cl2}\)

For 0.0800 mol of \(\ce{P4}\), the exact amount of chlorine required is:

\(\begin{aligned} n(\ce{Cl2\ required}) &=0.0800\times6 =0.480\text{ mol} \end{aligned}\)

That is exactly the amount present.

The coefficient comparison gives the same result:

\(\begin{aligned} \frac{0.0800}{1}&=0.0800 \\ \frac{0.480}{6}&=0.0800 \end{aligned}\)

Because the values are equal, the reactants are present in the exact stoichiometric ratio. Under the ideal model, neither remains in excess.

The amount of product is:

\(\begin{aligned} n(\ce{PCl3}) &=0.0800\times\frac{4}{1} =0.320\text{ mol} \end{aligned}\)

The misconception is treating a large coefficient as evidence that a reactant must be limiting. Coefficients describe the required ratio. They don’t tell you whether the required amount was actually supplied.

Question 5

Carbon monoxide reacts with hydrogen according to:

\(\ce{CO + 2H2 -> CH3OH}\)

A vessel initially contains 0.500 mol of \(\ce{CO}\) and 0.900 mol of \(\ce{H2}\).

After the reaction, an analysis reports that 0.040 mol of \(\ce{CO}\) remains.

A student concludes that hydrogen must have been the limiting reagent because carbon monoxide remained. Is that conclusion consistent with the ideal stoichiometric model in which the reaction proceeds until the limiting reagent is completely consumed? Justify your answer quantitatively.

Solution 5

Yes. The observation is consistent with hydrogen being the limiting reagent, and the ideal model predicts 0.050 mol of \(\ce{CO}\) remaining, which is close to but not exactly the reported 0.040 mol.

First identify the limiting reagent from the initial quantities.

For:

\(\ce{CO + 2H2 -> CH3OH}\)

compare:

\(\begin{aligned} \frac{0.500}{1}&=0.500 \\ \frac{0.900}{2}&=0.450 \end{aligned}\)

Hydrogen has the smaller value, so \(\ce{H2}\) is limiting.

If all 0.900 mol of hydrogen reacts, the carbon monoxide consumed should be:

\(\begin{aligned} n(\ce{CO\ reacted}) &=0.900\times\frac{1}{2} =0.450\text{ mol} \end{aligned}\)

The predicted amount remaining is therefore:

\(\begin{aligned} n(\ce{CO\ remaining}) &=0.500-0.450 =0.050\text{ mol} \end{aligned}\)

The reported value is 0.040 mol, so it differs from the ideal prediction by:

\(0.050-0.040=0.010\text{ mol}\)

The student’s qualitative conclusion, that hydrogen is limiting because carbon monoxide remains, agrees with the initial stoichiometric calculation. However, the measured amount does not match the simple ideal model exactly.

That difference could indicate experimental uncertainty, measurement error, an additional process consuming some \(\ce{CO}\), or that one of the stated quantities is inaccurate. Stoichiometry tells us what the ideal reaction predicts. Experimental evidence then lets us test whether that model fully describes what happened.

13Where limiting reagent calculations lead next

Once you can identify the limiting reagent, you can separate two different questions that are often mixed together.

First, how much product should be possible? That is the theoretical yield, calculated from the limiting reagent.

Second, how much product was actually obtained? Comparing the experimental amount with the theoretical amount leads directly to percentage yield:

\(\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times100\%\)

Limiting reagent calculations therefore give you the benchmark. Percentage yield tells you how closely the real experiment reached it.