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In this article, the author has explained what is a limiting reactant, definition, examples and method to find a limiting reactant in during a chemical reaction.
Limiting reactant Definition
Table of Contents
During most of the chemical reactions two or more reactants participate in product formation. Depending upon the nature of reactants and reaction conditions one of the reactants consumes earlier. Hence this is called a limiting reactant.
This reactant is much important while dealing with a chemical reaction.
The reactant that controls the amount of products formed in a chemical reaction and is consumed earlier is called a limiting reactant.
Watch the video lecture to better understand about limiting reactant
Explanation of limiting reactant
If stoichiometry amounts are put in chemical reactions, these amounts are completely used. Sometimes, stoichiometric amounts required by the reaction are not used during the reaction for the following purposes:
- To ensure that whole of the expensive reactant is completely consumed
- To make the reaction occur faster
- To complete 100% reaction
- To measure the yield of a balance chemical reaction
When we use non-stoichiometric amounts, one of the reactants is consumed earlier than the other. The reactant which is used in less amount is of products formed and is called a limiting reactant.
Related: Avogadro’s number and Molar Volume-Themasterchemistry.com
How to find limiting reactant
To determine the limiting reactant following steps should be taken
- Get the number of moles of reacting substances from the given amounts of reactants.
- Calculate the number of moles of the required product from molar amounts of all reactants.
- Choose the reactant that produces the least amount of product. This is known as a limiting reactant.
Examples to understand a limiting reactant
Example 1
In a chemical reaction, a large quantity of oxygen makes things burn rapidly. Since oxygen used in excess is left behind when the reaction is completed. The other reagent is consumed completely. That reactant that is consumed earlier is known as limiting reactant.
Example 2
The concept of limiting reactant is quite similar to the relation between 30 eggs and 58 slices to make the sandwiches.
As 1 egg and 2 slices make 1 sandwich. Therefore, only 29 sandwiches can be prepared from 58 slices.
One egg is left behind. 58 slices are completely used which limits the product sandwiches. So slices are limiting reactants.
Example 3
Consider the reaction along with stoichiometric amounts
2H2 + O2 ———-> 2H2O
(2 moles) (1 mole) (2 moles)
4g 32g 36g
If we use stoichiometric amounts then 4g amounts of hydrogen and 32 g of oxygen, we get 36g of water.
But if we take non-stoichiometric amounts say 4g of H2 and 60g of oxygen, we will not get more quantity of water than 36g because H2 is less. So H2 is a limiting reactant. 28 g of oxygen will be left unreacted.
2H2 + O2 ——-> 2H2O
4g 36 g 36g
Limiting reactant problems
Problem 1
NH3 gas can be prepared by heating together two solids, NH4Cl and Ca(OH)2. If a mixture containing 100 g of each solid is heated, then
a) Calculate the number of grams of NH3 produced
b) Calculate the excess amount of reagent left unreacted
Solution
The balanced chemical equation for the chemical reaction is
2NH4Cl + Ca(OH)2 —-> CaCl2 + 2NH3 + 2H2O
a) Number of grams of NH3 produced
given mass of NH4Cl = 100g
Molar mass of NH4Cl = 14+1×4+35.5 = 53.5 g/mol
Number of moles of NH4Cl= 100/53.5 = 1.87 moles
Given mass of Ca(OH)2 = 100 g
Molar mass of Ca(OH)2 = 40 + 2(16+1) = 74 g
Number of moles of Ca(OH)2 = 100/74 = 1.35 moles
–> Compare NH4Cl and NH3 according to a balanced chemical equation.
NH4Cl : NH3
2 moles : 2 moles
Therefore 1.87 moles: 2/2 x 1.87 = 1.87 moles
–> Compare Ca(OH)2 and NH3 according to the balanced chemical equation
Ca(OH)2 : NH3
2 moles : 2 moles
Therefore, 1.35 moles: 2 x 1.35= 2.70 moles
Since NH4Cl produces the least amount of NH3, hence NH4Cl is the limiting reactant.
Thus no. of moles of NH3 produced= 1.87 moles
Molar mass of NH3 = 14+1 x3 = 17 g/mol
Hence the amount of NH3 produced = No.of moles x Molar mass of NH3 as given below.
Amount of NH3 produced = 1.87 x 17 = 31.79 g
b) Amount of reagent left unreacted
–> Compare NH4Cl and Ca(OH)2 according to a balanced chemical equation.
NH4Cl : Ca(OH)2
2 moles : 1 mole
1.87 moles : 1×1.87/2 = 0.935 moles
Moles of Ca(OH)2 taken = 1.35 moles
Therefore unreacted moles = 1.35 – 0.935 = 0.415 moles
Thus mass of Ca(OH)2 left = 0.415 x 74 = 30.71 g
Problem 2
Calculate the number of grams of Al2S3, which can be prepared by the reaction of 20 g of Al and 20g of sulfur. How much non-limiting reactant is in excess?
Solution
The balanced chemical equation for the chemical reaction is
2Al + 3S —–> Al2S3
Given mass of Al = 20 g
Number of moles of Al = 20/27 = 0.74 moles
Given mass of S = 30 g
Number of moles of S = 30/32 = 0.9375 moles
Determination of limiting reactant in this reaction
Compare Al and Al2S3 according to a balanced chemical equation.
Al : Al2S3
2 moles : 1 mole
Therefore, 0.74 moles: ½ x 0.74 = 0.37 moles
Compare the number of moles of the product produced by Al = 0.37
–> Compare S and Al2S3 according to a balanced chemical equation.
S : Al2S3
3 moles : 1 mole
Therefore, 0.9375 moles: 1/3 x 0.9375 = 0.3215 moles
Hence the number of moles of Al2S3 produced by S = 0.3225 moles
Since S produces the least number of moles of the product, therefore it is the limiting reactant.
Hence moles of Al2S3 produced = 0.3125 moles
Mass of Al2S3 produced = 0.3125 x 150 = 46.87 g
Determination of amount of Al left unreacted
2Al + 3S ——> Al2S3
–> Compare the moles of S and Al to find the moles of Al reacted
S : Al
3 moles : 2 moles
0.9375 moles: 2/3 x 0.9375 = 0.625 moles
Moles of Al consumed = 0.625 moles
Moles of Al taken = 0.74 moles
Moles of Al left unreacted = 0.74 – 0.625 = 0.115 moles
Mass of Al left unreacted: 0.115 x 27 = 3.105 g
Related questions
1: How does a limiting reactant help to control the reaction?
As we know a limiting reactant is the one that is in a lower amount and is consumed earlier. After complete consumption of limiting reactant further formation of product stops and excess reagent is left in the reaction media. Therefore, if the limiting reagent is not available to the excess reagent then the product cannot be formed further. In this way, a limiting reagent controls the chemical reaction.
2: 11g of carbon is reacted with 32 g of oxygen to give CO2. Which is the limiting reactant?
C + O2 ——> CO2
According to a balanced chemical equation, there should be 44 grams of CO2 after the complete reaction of 12g of carbon with 32 grams of oxygen. Therefore, 32 grams of oxygen will be in excess compared to 11 grams of carbon. So in this case carbon is a limiting reactant.
3: Why the concept of limiting reactants is not applicable to reversible reactions?
For a reaction having one of the reactants as a limiting one, one of the reactants has to be consumed completely. But in the case of a reversible reaction, at the stage of equilibrium, certain amounts of reactants are left behind. Therefore, the reversible reactions do not give any idea about the concept of limiting reactants.
4. How do you find the limiting reactant?
You need to perform a titration analysis to determine the limiting reactant. For example, if you have two reactants, A and B, then you can determine the limiting reactant by using a stoichiometric equation, such as (A + B) = C.
5. What is a limiting and excess reactant?
Reactants are substances that react to form new compounds. Limiting reactants are the ones that cannot be combined to form a compound with a greater quantity of the element. Excess reactants are those that are left after you’ve added the limiting reactants to your reaction. They can be used again in future reactions.
6. What is an excess reactant?
An excess reactant is a substance that is left over after the reaction of a chemical equation has taken place. An example of this would be when you mix a cup of salt and sugar to create a rock candy, and you end up with about 2 cups of salt and 1/2 cup of sugar. In this case, the remaining 1/2 cup of sugar is the excess reactant.