# Limiting Reagent

In a chemical reaction, the limiting reagent is called as the reactant which determines the quantity of the products that are made. The other reactants present in the reactions are sometimes called as being in excess since there is some leftover quantity of them after the limiting reagent is completely used up. The maximum amount of product that is produced is known as the theoretical yield. The limiting reagent should be identified to calculate the percentage yield of a reaction. Given the balanced chemical equation, that describes the reaction, there are many equivalent ways to identify the limiting reagent and calculate the excess quantities of other reagents in the reaction. In this article, we learn about what is limiting agent, how to find limiting reagents and some limiting reagent questions.

### Limiting Reagent Definition

Limiting reagents are defined as the substances which are entirely consumed in the completion of a chemical reaction. They are also referred to as limiting reactants or limiting agents. According to the stoichiometry of chemical reactions, a fixed amount of reactants is necessary for the reaction to complete.

This reactant usually determines when the reaction would stop. The exact amount of reactant that would be needed to react with another element is calculated from the reaction stoichiometry. The limiting reagent depends on the mole ratio and not on the masses of the reactants present.

Consider the following reaction for the formation of ammonia:

3H2 + N2 ---> 2NH3

In the reaction shown above, 3 moles of the hydrogen gas is required for the reaction with 1 mole of nitrogen gas for the formation of 2 moles of ammonia. But what if, during the time of the reaction, there are only 2 moles of hydrogen gas available with 1 mole of nitrogen?

In this case, the entire quantity of nitrogen cannot be used since the entirety of nitrogen requires 3 moles of hydrogen gas to react. Therefore, the hydrogen gas is limiting the reaction and is hence called as the limiting reagent for this reaction.

### Limiting Reagent Examples

Let us now look at some of the limiting reagent examples.

Example

Consider the combustion of benzene which is represented by the following chemical equation:

2C6H6(l) + 15 O2(g) ---> 12CO2(g) + 6HO2(l)

It means that 15 moles of molecular oxygen O2 is needed to react with 2 moles of benzene C6H6

The amount of oxygen that is required for other quantities of benzene is calculated using cross-multiplication. For example, if 1.5 mol C6H6 is present, 11.25 mol O2 is required:

1.5 mol C6H6 x $\frac{15\;mol\;O_{2}}{2\;mol\;C_{6}H_{6}}$ = 11.25 mol O2

If in 18 mol O2 are present, there would be an excess of (18 - 11.25) = 6.75 mol of unreacted oxygen when all of the benzene is consumed. Benzene is, therefore, the limiting reagent.

### How to Find Limiting Reagent in a Reaction?

Let us now learn about how to determine limiting reagent in a reaction.

There are two ways for how to calculate limiting reagent. One method is to find and compare the mole ratio of the reactants that are used in the reaction. Another method is to calculate the grams of products produced from the quantities of reactants in which the reactant which produces the smallest amount of product is the limiting reagent.

Method 1: Finding the limiting reagent by looking at the number of moles of every reactant.

1. First, determine the balanced chemical equation for the given chemical reaction.

2. Then, convert all the given information into moles (by using molar mass as a conversion factor).

3. The next step is to calculate the mole ratio from the given information. Then, compare the calculated ratio to the actual ratio.

4. Use the amount of limiting reactant for calculating the amount of product produced.

5. Lastly, if necessary, calculate how much of the non-limiting agent is left in excess.

Method 2: Finding the limiting reagent by calculating and comparing the amount of product each reactant would produce.

1. The first step is to balance the chemical equation for the given chemical reaction.

2. Then, convert the given information into moles.

3. Use stoichiometry for each individual reactant for finding the mass of product produced.

4. The reactant which produces a lesser amount of product would be the limiting reagent.

5. The reactant which produces a larger amount of product would be the excess reagent.

6. Lastly, for finding the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass given of the excess reagent.

### Limiting Reagent Problems

1. Determine the limiting reagent if 76.4 grams of  C2H3Br3 reacts with 49.1 grams of O2.

4 C2H3Br3 + 11 O2 ---> 8 CO2 + 6HO2 + 6Br2

Solution:

Using method 1,

76.4 g x  $\frac{1\;mol}{266.72g}$ = 0.286 moles of C2H3Br3

49.1 g x  $\frac{1\;mol}{32g}$ = 1.53 moles of O2

If you assume that all of the oxygen is used up,

1.53 x 411, or, 0.556 moles of C2H3Br3 are required. Since there are only 0.286 moles of C2H3Br3 that are available, C2H3Br3 is the limiting reagent here.

Using method 2,

76.4 g C2H3Br3 $\frac{1\;mol}{266.72g}$ x  $\frac{8\;mol\;CO_{2}}{4\;mol\;C_{2}H_{3}Br_{3}}$ x $\frac{44.01\;g\;CO_{2}}{1\;mol\;CO_{2}}$ = 25.2 g CO2

49.1 g O2 x $\frac{1\;mol\;O_{2}}{32\;g\;O_{2}}$ x $\frac{8\;mol\;CO_{2}}{11\;mol\;O_{2}}$ x $\frac{44.01\;g\;CO_{2}}{1\;mol\;CO_{2}}$ =  49.1 g CO2

Hence, by using any of these methods, C2H3Br3 is the limiting reagent.