# Limiting Reactant A limiting reactant is completely used up in a reaction, and other reactants are excess reactants. The amount of product is always calculated based on the number of moles of the limiting reactant.

When a reaction is carried out in a laboratory, the reactants may not always be present in the perfect stoichiometric ratio, as shown by the balanced equation. A system may have an excess of one reactant or the other. The reactant that is completely used up in a reaction, which determines the amount of product that can form, is the limiting reactant (or limiting reagent). The reactant that is not completely used up in a reaction is the excess reactant (or excess reagent).

The steps to identify the limiting reactant are:

1. Determine the number of moles of the first reactant, and calculate the moles of the product that could form from it.
2. Determine the number of moles of the second reactant, and calculate the number of moles of the product that could form from it.
3. Compare the number of moles of the product that each reactant can form. The reactant that produces fewer moles of product is the limiting reactant.
4. Calculate the theoretical yield from the number of moles of product the limiting reactant can produce.
5. To determine the amount of excess reactant that will remain after the reaction is complete, compare the ratios of the limiting and excess reactants, calculate the moles of excess reactant used to convert to mass. Subtract the value from the initial mass of the excess reactant.

Step-By-Step Example
Identifying the Limiting Reactant in the Combustion of Ammonia
A mass of 4.50 g of ammonia (NH3) reacts with 5.20 grams (g) of oxygen (O2). Which is the limiting reactant, and which is the excess reactant? What mass of nitric oxide (NO) does the reaction form? What amount of excess reactant will remain after the reaction is complete?
Step 1
Write the balanced equation.
$4{\rm{NH}}_3+5{\rm O}_2\;\rightarrow4\rm{NO}+6{\rm H}_2\rm O$
Step 2
Calculate the molar mass of NH3. Multiply the number of each atom in the formula by its molar mass.
\begin{aligned}\text{Molar mass of }{\rm NH_{3}}&=(1)(14.01\rm{\; g/mol})+(3)(1.01\rm{\; g/mol})\\&=17.04\rm{\; g/mol}\end{aligned}
Step 3
Use the molar mass of NH3 to convert the given mass to the number of moles.
\begin{aligned}\text{Moles of }{\rm NH_{3}}&=\frac{4.50\rm{\; g}}{17.04\rm{\; g/mol}}\\&=0.2641\rm{\; mol\; NH}_3\end{aligned}
To three significant figures, there are 0.264 mol NH3.
Step 4
The balanced equation shows that the mole ratio of NH3 to NO is 4:4. Therefore, 4.50 g of NH3 can produce 0.2641 mol NO.
Step 5
Use the molar mass of O2 to convert the given mass to number of moles.
\begin{aligned}\text{Moles of }{\rm O_{2}}&=\frac{5.20\rm{\; g}}{32.00\rm{\; g/mol}}\\&=0.1625\rm{\; mol\; O}_2\end{aligned}
Step 6
The balanced equation shows that the mole ratio of O2 to NO is 5:4. Use dimensional analysis to convert moles of O2 to moles of NO using the mole ratio.
\begin{aligned}\text{Moles of }{\rm NO}&=(0.1625\rm{\; mol\; O}_2)\!\left(\frac{4\rm{\; mol\; NO}}{5\rm{\; mol\; O}_2}\right)\\&=0.1300\rm{\; mol\; NO}\end{aligned}
To three significant figures, there are 0.130 mol NO.
Step 7
The given mass of NH3 can produce 0.2641 mol NO, and the given mass of O2 can produce 0.1300 mol NO. Therefore O2 is is the limiting reactant because it produces the least amount of product.
Step 8
Calculate the molar mass of NO. Multiply the number of each atom in the formula by its molar mass.
\begin{aligned}\text{Molar mass of }{\rm NO}&=(1)(14.01\rm{\; g/mol})+(1)(16.00\rm{\; g/mol})\\&=30.01\rm{\; g/mol}\end{aligned}
Step 9
Use the molar mass of NO and the number of moles of NO that can be produced by O2 to calculate the mass of NO that is produced.
\begin{aligned}\text{Mass of }{\rm NO}&=(0.1300\rm{\; mol\; NO})(30.01\rm{\; g/mol})\\&=3.90\rm{\; g\; NO}\end{aligned}
Step 10
To calculate the number of moles of excess reactant after the reaction is complete, notice that the balanced equation shows that the mole ratio of NH3 to O2 is 4:5. Use the mole ratio to calculate the moles of NH3 that will be used in the reaction.
\begin{aligned}\text{Moles of }{\rm NH_{3}}\text{ used}&=(0.1625\rm{\; mol\; O}_2)\!\left(\frac{4\rm{\; mol\; NH}_3}{5\rm{\; mol\; O}_2}\right)\\&=0.1300\rm{\; mol\; NH}_3\end{aligned}
Subtract this amount from the initial amount of NH3 (0.2641 mol) to identify the number of moles of excess NH3.
\begin{aligned}\text{Moles of excess }{\rm NH_{3}}&=0.2641\rm{\; mol}-0.1300\rm{\; mol}\\&=0.1341\rm{\; mol\; NH}_3\end{aligned}
To three significant figures, there are 0.134 mol NH3.
Step 11
Use the molar mass of NH3 to convert the moles to mass.
\begin{aligned}\text{Mass of excess }{\rm NH_3}&=(0.1341\rm{\; mol})(17.04\rm{\; g/mol})\\&=2.29\rm{\; g}\end{aligned}
Solution
The limiting reactant is O2, and the excess reactant is NH3. The reaction produces 3.90 g NO with an excess of 2.29 g NH3 remaining.
Step-By-Step Example
Identifying the Limiting Reactant in the Reaction of Copper(II) Oxide and Sulfuric Acid
A reaction beginning with 8.00 grams (g) of copper(II) oxide (CuO) and a solution of 7.10 g of sulfuric acid (H2SO4) produces copper sulfate (CuSO4) and water. Which is the limiting reactant, and which is the excess reactant? What mass of copper sulfate does the reaction form? How much of the excess reactant remains after the reaction is complete?
Step 1
Write the balanced equation.
$\rm{CuO}+{\rm H}_2\rm{SO}_4\rightarrow\rm{CuSO}_4+{\rm H}_2\rm O$
Step 2
Calculate the molar mass of CuO. Multiply the number of each atom in the formula by its molar mass.
\begin{aligned}\text{Molar mass of }{\rm CuO}&=(1)(63.55\rm{\; g/mol})+(1)(16.00\rm{\; g/mol})\\&=79.55\rm{\; g/mol}\end{aligned}
Step 3
Use the molar mass of CuO to convert the mass to number of moles.
\begin{aligned}\text{Moles of }{\rm CuO}&=\frac{8.00\rm{\; g}}{79.55\rm{\; g/mol}}\\&=0.1006\rm{\; mol\; CuO}\end{aligned}
Step 4
The balanced equation shows that the mole ratio of CuO to CuSO4 is 1:1. Therefore, 8.00 g of CuO can produce 0.1066 mol CuSO4.
Step 5
Calculate the molar mass of H2SO4. Multiply the number of each atom in the formula by its molar mass.
\begin{aligned}\text{Molar mass of }{\rm H_{2}SO_{4}}&=(2)(1.01\rm{\; g/mol})+(1)(32.06\rm{\; g/mol})+(4)(16.00\rm{\; g/mol})\\&=98.08\rm{\; g/mol}\end{aligned}
Step 6
Use the molar mass of H2SO4 to convert the mass to number of moles.
\begin{aligned}\text{Moles of }{\rm H_{2}SO_{4}}&=\frac{7.10\rm{\; g}}{98.08\rm{\; g/mol}}\\&=0.07239\rm{\; mol\; H}_2\rm{SO}_4\end{aligned}
Step 7
The balanced equation shows that the mole ratio of H2SO4 to CuSO4 is 1:1. Therefore 7.10 g H2SO4 can produce 0.07239 mol CuSO4.
Step 8
The given mass of CuO can produce 0.1066 mol CuSO4, and the given mass of H2SO4 can produce 0.07239 mol CuSO4. Therefore H2SO4 is the limiting reactant because it produces the least amount of product.
Step 9
Calculate the molar mass of CuSO4. Multiply the number of each atom in the formula by its molar mass.
\begin{aligned}\text{Molar mass of }{\rm CuSO_{4}}&=(1)(63.55\rm{\; g/mol})+(1)(32.06\rm{\; g/mol})+(4)(16.00\rm{\; g/mol})\\&=159.61\rm{\; g/mol}\end{aligned}
Step 10
Use the molar mass of CuSO4 and the number of moles of CuSO4 that can be produced by H2SO4 to calculate the mass of CuSO4 that is produced.
\begin{aligned}\text{Mass of }{\rm CuSO_{4}}&=(0.07239\rm{\; mol\; CuSO}_4)(159.61\rm{\; g/mol})\\&=11.6\rm{\; g\; CuSO}_4\end{aligned}
Step 11

To calculate the number of moles of excess reactant after the reaction is complete, notice that the balanced equation shows that the mole ratio of CuO to H2SO4 is 1:1. Therefore, 0.07239 mol CuO is used in the reaction.

Subtract this amount from the initial amount of CuO (0.1066 mol) to determine the number of moles of excess CuO.
\begin{aligned}\text{Moles of excess }{\rm CuO}&=0.1066\rm{\; mol}-0.07239\rm{\; mol}\\&=0.03421\rm{\; mol\; CuO}\end{aligned}
Step 12
Use the molar mass of CuO to convert the moles to mass.
\begin{aligned}\text{Mass of excess }{\rm CuO}&=(0.03421\rm{\; mol})(79.55\rm{\; g/mol})\\&=2.72\rm{\; g}\end{aligned}
Solution
The limiting reactant is H2SO4, and the excess reactant is CuO. The reaction produces 11.5 g CuSO4 with an excess of 2.72 g CuO remaining.