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# Limiting Reactant

Top
 Sub Topics Most of the chemical reactions the reactant will not be used in stoichiometric proportion or quantities. One of the reactants will be present in excess and remain unreacted even when the other reactant has completely reacted. In a chemical reaction suppose if the extent of reaction is not limited by thermodynamic equilibrium constraints, this limiting reagent is the one that determines the maximum possible value of the extent of reaction. An analogous situation occurs in chemical reactions when one reactant is used up before the others. The reaction stops as soon as any reactant is totally consumed, leaving the excess reactants as leftovers.

## Definition

"The reactant which present in excess is termed excess reactant and the other reactant which is present in a lesser quantity and cannot react with whole of the other reactant is called limiting reactant."

The amount by which any reactant is present present in excess to that required to combine with the limiting reactant is usually expressed as percentage excess. All calculations involved in estimating the quantity of products and conversion are always based on the limiting reactant. A limiting reactant is the one which will not be present in the product whereas the excess reactant is the one which will always be present in the product.

## Stoichiometry Limiting Reactant

The word stoichiometry refers to quantitative aspects of chemistry and limiting reactant usually involve determining the amount of product from a given amount of two or more reactants. Stoichiometry is an area that usually involves quantitative manipulations of chemical quantities such as grams and moles. A limiting reactant simply is the first reactant in a balanced chemical equation that will fully transform or react. Because the limiting reactant will be the first to entirely change, it controls how much of the product or products will be made.

The problems that involve this relationship are commonly known as mass-mass problems. An example is shown below.

### Solved Example

Question: Using the equation, N2 + 3H2 $\rightarrow$ 2NH3, how many grams of NH3 is produced from 25.0 grams of nitrogen and 5.0 grams of hydrogen gases?
Solution:
Solution using dimensional analysis

Step-1: $25.0g\ N_{2} \times$ $\frac{1\ mol\ N_2}{28.0g\ N_2}$ $\times$ $\frac{2\ mol\ NH_3}{1\ mol\ N_2}$ $\times$ $\frac{17.0g\ NH_{3}}{1\ mol\ NH_{3}}$ = 30.4g NH3

Step-2: $5.0g\ H_2 \times$ $\frac{1\ mol\ H_2}{2.016g\ H_2}$ $\times$ $\frac{2\ mol\ NH_{3}}{3\ mol\ H_2}$ $\times$ $\frac{17.0g\ NH_{3}}{1\ mol\ NH_{3}}$ = 28.1g NH3

## Why have Limiting Reactants?

A limiting reactant is the reactant that is completely converted to products during the reaction. Once the limiting reactant has been use up, no more product can form. The limiting reactant must be used as the basis for calculating the maximum possible amount of products because the limiting reactant limits the amount of products that can be formed. In chemical reactions the yield always depends on the amount of the limiting reactant. The reactant that gives less product is the limiting reactant.

The molecular view of formation of water is shown below.

When a equal number of moles of H2 and O2 are reacted according to the equation

2H2 + O2 $\rightarrow$ 2H2O

All H2 completely reacts whereas only half of the O2 is consumed. In this case, the H2 is the limiting reactant and the O2 is the excess reactant.

## How to Find Limiting Reactant with Steps?

Limiting reactant problems are more complicated when the quantities are expressed in measurable mass units. the overall procedure to find the liming reactant is given below.
1. Convert the number of grams of each reactant to moles.
2. Identify the limiting reactant.
3. Calculate the number of moles of each species that reacts or is produced.
4. Calculate the number of moles of each species that remains after the reaction.
5. Change the number of moles of each species to grams.

## Method for Determining Limiting Reactant

A limiting reactant is the reactant totally consumed in a reaction, thereby determining the maximum yield possible.

The smaller amount method for determining limiting reactant is
• Calculate the amount of product that can be formed by the initial amount of each reactant.
1. The reactant that yields the smaller amount of product is the limiting reactant.
2. The smaller amount of product is the amount that will be formed when all of the limiting reactant is used up.
• Calculate the amount of excess reactant that is used by the total amount of limiting reactant.
• Subtract from the amount of excess reactant present initially the amount that is used by all of the limiting reactant, the difference is the amount of excess reactant that is left.

## The Yield Concept

The amount of products calculated theoretically called theoretical yields determined by assuming that the reaction goes cleanly and completely. The actual yield of a product (that is the amount present after separating it from other products and reactants and purifying it) is less than the theoretical yield. There are several reasons for this.

The reaction may stop short of completion, so reactants remain un reacted. There may be comparing reactions that give other products, and therefore reduce the yield of the desired one. Finally in the process of separation and purification, some of the product is invariably lost, although that amount can be reduced by careful experimental techniques. The ratio of the actual yield to the theoretical yield gives the percentage yield for that product in the reaction.

## Lab Experiment Procedure

To determine the limiting reactant in a mixture of two soluble salts, the following techniques are used in the experimental procedure.

The reactant determining the amount of product generated in a chemical reaction is called the limiting reactant in the chemical system. To determine the limiting reactant of the two given reactants, calculate the theoretical yield of the product and to find how much of the excess reactant is unused.

In the laboratory , it is often more convenient to have one or more reactant present in excess, so that not all the reactants are present in stoichiometric amounts. However, it is the reaction stoichiometry that determines how much product is formed.

The limiting reactant is the substance in a chemical reaction that runs out first. When the limiting reactant is used up the reaction stops. At the end of a reaction the limiting reactant will be completely consumed and one or more other reactants will be left over in the reaction mixture. These reactants are known as excess reactants.

## Examples

Let us consider that 18 kg of carbon is burnt with 32 kg of oxygen. As per stoichiometry

C + O2 $\rightarrow$ CO2

That is 12 kg of carbon will burn with 32 kg of oxygen to form 44 kg of CO2.

Hence for 18 kg of carbon to react fully we should have 48 kg of oxygen. Since 32 kg of oxygen alone is available, it is called the limiting reactant and carbon is called the excess reactant. For 32 kg of oxygen to react fully, it is sufficient to have 12 kg of carbon. However 6 kg of carbon is present in excess.

Hence % excess of carbon is
$\frac{6}{12}$ $\times 100$ = 50%

## Problems

Some of the problems based on limiting reactant is given below.

### Solved Examples

Question 1: Zinc metal reacts with hydrochloric acid bu the following reaction.
Zn(s) + 2HCl(aq) $\rightarrow$ ZnCl2(aq) + H2(g)
If 0.30 mol Zn is added to hydrochloric acid containing 0.52 mol HCl, how many moles of H2 are produced?
Solution:
0.30 mol Zn $\times$ $\frac{1\ mol\ H_{2}}{1\ mol\ Zn}$ = 0.30 mol H2

0.50 mol HCl $\times$ $\frac{1\ mol\ H_{2}}{2\ mol\ HCl}$ = 0.26 mol H2

Hydrochloric acid must be the limiting reactant and that some zinc must be left unconsumed. (Zinc is the excess reactant)

Since HCl is the limiting reactant the amount of H2 produced must be 0.26 mol.

Question 2: If 28 g of Fe react with 24 g of S to produce FeS, what would be the limiting reactant? How many grams of excess reactant would be present in the vessel at the end of the reaction?

Fe + S $\overset{\Delta }{\rightarrow}$ FeS
Solution:
The number of moles for each reactant must be determined first.

28 g Fe $\times$ $\frac{1\ mol\ Fe}{56g}$ = 0.5 mol Fe

24 g S $\times$ $\frac{1\ mol\ S}{32g}$ = 0.75 mol S

Since one mole of Fe is needed to react with one mole of S, and there are 0.5 moles Fe for every 0.75 moles S, the limiting reagent is Fe. Thus 0.5 moles of Fe will react with 0.5 moles of S, leaving an excess of 0.25 moles of S in the vessel. The mass of the excess reactant will be

mass of S = (0.25 mol S) $\frac{32g}{1\ mol\ S}$

= 8 g of S