How to Find the Limiting Reactant (Step by Step)

Think of making sandwiches. Each sandwich needs 2 slices of bread and 1 slice of cheese. With 10 slices of bread and 3 slices of cheese, you can only make 3 sandwiches, because the cheese runs out first. The cheese is the limiting reactant, and the 4 extra slices of bread are the excess.

A reaction works the same way, except the "recipe" is the balanced equation and the amounts must be in moles. Grams alone can fool you, because different substances have very different molar masses.

The full process:

1. Balance the equation.
2. Convert every reactant amount to moles.
3. Use the mole ratio to see which reactant runs out first.
4. Use the limiting reactant to calculate the product.

This is the most reliable method because it also gives you the theoretical yield.

Problem: 10.0 g of aluminum reacts with 35.0 g of chlorine gas. Which is limiting, and how many grams of $\mathrm{AlCl_3}$ can form?

$$\mathrm{2Al + 3Cl_2 \rightarrow 2AlCl_3}$$

Molar masses (g/mol): Al 26.98, $\mathrm{Cl_2}$ 70.90, $\mathrm{AlCl_3}$ 133.33.

1. Convert both reactants to moles.
   • Al: $10.0 \div 26.98 = 0.3706$ mol
   • $\mathrm{Cl_2}$: $35.0 \div 70.90 = 0.4937$ mol
2. Find the product each one could make.
   • From Al: $0.3706 \text{ mol Al} \times \frac{2 \text{ mol AlCl}_3}{2 \text{ mol Al}} = 0.3706$ mol $\mathrm{AlCl_3}$
   • From $\mathrm{Cl_2}$: $0.4937 \text{ mol Cl}_2 \times \frac{2 \text{ mol AlCl}_3}{3 \text{ mol Cl}_2} = 0.3291$ mol $\mathrm{AlCl_3}$
3. The smaller amount wins. $\mathrm{Cl_2}$ makes less product, so $\mathrm{Cl_2}$ is the limiting reactant.
4. Convert to grams:

$$0.3291 \text{ mol} \times 133.33 \text{ g/mol} = 43.9 \text{ g AlCl}_3$$

Answer: $\mathrm{Cl_2}$ is limiting, and the theoretical yield is 43.9 g of $\mathrm{AlCl_3}$. Notice that $\mathrm{Cl_2}$ had the larger mass but is still the limiting reactant.

A faster check is to divide each reactant's moles by its coefficient in the balanced equation. The smallest result is the limiting reactant.

Using the same problem:

1. Al: $0.3706 \div 2 = 0.1853$
2. $\mathrm{Cl_2}$: $0.4937 \div 3 = 0.1646$
3. $0.1646 < 0.1853$, so $\mathrm{Cl_2}$ is limiting.

This tells you which reactant is limiting, but you still need one more step (Method 1, step 4) to get the mass of product. The numbers from this shortcut are not moles of anything real, so do not use them as your final answer.

You can also pick one reactant and ask how much of the other it needs.

Problem: 28.0 g of $\mathrm{N_2}$ reacts with 8.00 g of $\mathrm{H_2}$. How much ammonia forms?

$$\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$$

Molar masses (g/mol): $\mathrm{N_2}$ 28.02, $\mathrm{H_2}$ 2.016, $\mathrm{NH_3}$ 17.03.

1. Moles you have: $\mathrm{N_2}$: $28.0 \div 28.02 = 0.9993$ mol. $\mathrm{H_2}$: $8.00 \div 2.016 = 3.968$ mol.
2. Moles of $\mathrm{H_2}$ needed to use up all the $\mathrm{N_2}$: $0.9993 \times \frac{3}{1} = 2.998$ mol.
3. Compare: you need 2.998 mol of $\mathrm{H_2}$ and have 3.968 mol. There is more than enough $\mathrm{H_2}$, so $\mathrm{N_2}$ is the limiting reactant.
4. Product from the limiting reactant: $0.9993 \text{ mol N}_2 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol N}_2} = 1.999$ mol, and $1.999 \times 17.03 = 34.0$ g of $\mathrm{NH_3}$.

If the amount needed had been larger than the amount you have, the other reactant would be limiting instead.

Once you know the limiting reactant, you can find how much of the excess reactant is left.

1. Use the limiting reactant to find how much excess reactant is used.
2. Subtract that from the amount you started with.

Aluminum example: $\mathrm{Cl_2}$ is limiting with 0.4937 mol.

1. Al used: $0.4937 \text{ mol Cl}_2 \times \frac{2 \text{ mol Al}}{3 \text{ mol Cl}_2} = 0.3291$ mol Al.
2. In grams: $0.3291 \times 26.98 = 8.88$ g Al used.
3. Left over: $10.0 - 8.88 = 1.12$ g of Al.

Ammonia example: $\mathrm{H_2}$ left over is $3.968 - 2.998 = 0.970$ mol, which is $0.970 \times 2.016 = 1.96$ g.

Paste it into the VeilGPT solver to check your steps on a problem from your homework.