# Limiting Reactant or Limiting Reagent

In chemistry, the limiting reactant is the reactant that gets completely used up in a chemical reaction. It is also known as the limiting reagent, although a reactant and reagent are not always the same thing in modern chemistry. Here’s what you need to know about the limiting reactant, including its definition, how to find it, and worked example problems.

### Limiting Reactant Definition

The limiting reactant is the reactant that “limits” a chemical reaction or determines the amount of product that it can produce. It is based on stoichiometry or the mole ratio between reactants and products. When you combine reactants, you don’t always use amounts that perfectly balance each other out.

For example, if you buy a package of hot dogs and a package of buns, usually you get 10 hot dogs and 8 buns (presumably so bread companies can sell more of their product). How many hot dogs with buns can you make? The ideal ratio between hot dogs and buns is 1:1, but you have 10:8 or 1:0.8. You use up all the buns before you run out of hot dogs, so the buns are the limiting reactant. The leftover hot dogs are the excess reactant. The limiting reactant determines the theoretical yield. In this case, the theoretical yield is 8 hot dogs with buns.

Chemical reactions work a lot like the hot dog and buns example. For example, consider the reaction between iron (Fe) and sulfur (S) to form iron sulfide (FeS). Iron and sulfur, like hot dogs and buns, react in a 1:1 ratio:

Fe(s) + S(s) → FeS(s)

In an ideal situation, you have equal numbers of moles of each substance. You use a periodic table and look up the atomic masses of iron and sulfur and convert the numbers of moles to grams so you can weigh them out. So, if you react 55.85 grams of iron and 32.06 grams of sulfur, they react completely and form iron sulfide. Both iron and sulfur are the limiting reactant. If you use half the mass of each chemical or 5 times the amount of each chemical, the result is the same as long as iron and sulfur are present in equal molar amounts.

But, usually you have a given amount of each reactant, so one reactant gets used up (the limiting reactant) and there is some leftover reactant (the excess reactant). Let’s say you have 46 grams of iron and 32 grams of sulfur. Which is the limiting reactant?

### How to Find Limiting Reactant

There are two main ways of finding the limiting reactant. In both cases, you start with the balanced chemical equation and the number of moles of reactants and products. If you are given the number of moles, great! Usually, you have masses in grams and convert grams to moles. Then, you either compare the mole ratios of the reactants or you see which reactant produces the smaller amount of product:

#### Comparing the Mole Ratio of Reactants

1. Balance the chemical equation.
2. From the balanced equation, find the mole ratio between reactants. The coefficients in front of the reactants and products give you the mole ratio. If there is no coefficient, it means there is 1 mole.
3. Convert grams into moles.
4. Compare the number of moles of the actual amount of reactants to the mole ratio from the balanced equation.
5. The reactant that comes up short is the limiting reactant.

For example, in the reaction between 46 grams of iron and 32 grams of sulfur:

1. Write the balanced equation: Fe(s) + S(s) → FeS(s)
2. Find the mole ratio between reactants: 1 Fe: 1 S
3. Convert grams to moles: 46 grams Fe = 46 g / 55.85 g/mol = 0.82 mol; 23 grams S = 23 g / 32.06 g/mol = 0.72 mol
4. Compare actual ratio to ideal mole ratio: You have 0.82 mol Fe : 0.72 mol S
5. Which comes up short? Since Fe and S react in a 1:1 ratio and there are fewer moles of S, sulfur is the limiting reactant.

#### Comparing the Amount of Product

1. Balance the chemical equation.
2. Convert actual amounts into moles.
3. Using the mole ratio from the balanced equation, find the moles of product you get if you use the full amount of each product. So, if there are two reactants, there are two calculations here. If you have three reactants, then there are three calculations. If your reactions forms more than one product, just pick one.
4. The reactant that yields the smaller amount of product is the limiting reactant. The one that yields the larger amount of product is the excess reactant.

If you like, calculated the amount of excess reactant by subtracting the moles of excess reactant from the amount that actually gets used up. Then, convert the number of moles to grams.

Again, let’s use the example of the reaction between 46 grams of iron and 32 grams of sulfur.

1. Write the balanced equation: Fe(s) + S(s) → FeS(s) [Note that 1 mole of either reactant yields 1 mole of product.]
2. Convert grams to moles: 46 grams Fe = 46 g / 55.85 g/mol = 0.82 mol; 23 grams S = 23 g / 32.06 g/mol = 0.72 mol
3. Find theoretical moles of product from each reactant: Because the reactant:product ratio is 1:1, the maximum number of moles of product is 0.82 moles using all of the Fe or 0.72 moles of product using all of the S.
4. The amount of sulfur yields the small amount of iron sulfide, so sulfur is the limiting reactant.

### Example

Here is another example showing how to find limiting reactant:

Find the limiting reactant when you react 35.60 grams of sodium hydroxide and 30.80 grams of phosphoric acid to form sodium phosphate and water.

First, you need the chemical formulas and balanced chemical equation:

3 NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O(l)

The mole ratio between reactants is 3 NaOH: 1 H3PO4

Go ahead and find the number of moles of sodium hydroxide and phosphoric acid. Using the atomic masses from the periodic table:

• Molar mass of NaOH = 40.00 grams/mole
• Molar mass of H3PO4 = 98.00 grams/mol

moles of NaOH = 35.60 g / 40.00 g/mol = 0.89 mol
moles of H3PO4 = 30.80 g / 98.00 g/mol = 0.31 mol

The actual mole ratio between sodium hydroxide and phosphoric acid is 0.89 : 0.31 = 2.87.
The ideal mole ratio between sodium hydroxide and phosphoric acid is 3:1 = 3.00.

Since the mole ratio is less than 3, you know NaOH is the limiting reactant. If the ratio had been greater than 3, then phosphoric acid would have been the limiting reactant. Of course, you get the same answer working the problem by comparing moles of reactants to products.

### References

• Brady, James E.; Senese, Frederick; Jespersen, Neil D. (2007). Chemistry: Matter and its Changes. John Wiley & Sons. ISBN 9780470120941.
• Giunta, Carmen J. (2016). “What’s in a Name? Amount of Substance, Chemical Amount, and Stoichiometric Amount.” J. Chem. Educ. 93(4): 583-586. doi:10.1021/acs.jchemed.5b00690
• Olmsted, John; Williams, Gregory M. (1997). Chemistry: The Molecular Science. Jones & Bartlett Learning. ISBN 0815184506.
• Zumdahl, Steven S. (2006). Chemical Principles (4th ed.). New York: Houghton Mifflin Company. ISBN 0-618-37206-7.