Question

3 grams of ${{H}_{2}}$ reacts with 29 grams of ${{O}_{2}}$ to yield water
(1) Which is the limiting reagent?
(2) Calculate the maximum amount of water that can be formed.
(3) Calculate the amount of one of the reactants which remains unreacted?

Answer 1

The chemical reactant which is completely consumed with the completion of the chemical reaction is known as the limiting reagent or limiting reactant or limiting agent. The amount of product formed is governed by the limiting reagent, since the reaction cannot continue without it.

Complete step by step answer:
-We will begin the solution by writing the balanced equation for the question-
$2{{H}_{2}}+{{O}_{2}}\to 2{{H}_{2}}O$

(i) From the reaction, it is clear that 2 moles of ${{H}_{2}}$ reacts with 1 mole of ${{O}_{2}}$ to form 2 moles of ${{H}_{2}}O$.
-Calculating the molar mass of Hydrogen $=1\times 2=2g$
Calculating the molar mass of Oxygen$=16\times 2=32g$
-It is clear that 4g of ${{H}_{2}}$ reacts with 32 g of ${{O}_{2}}$.
Therefore, 3g of ${{H}_{2}}$ reacts with$=\dfrac{32}{4}\times 3g\text{ of }{{\text{O}}_{2}}=27g\text{ of }{{\text{O}}_{2}}$
-As the given amount of Oxygen is more than the required amount, therefore ${{O}_{2}}$ is an excess reagent, and ${{H}_{2}}$ is a limiting reagent.

-(ii) We can see that 2 moles of hydrogen gas react with 1 mole of Oxygen producing 2 moles of water.
4g of ${{H}_{2}}$ produces = 32g of ${{H}_{2}}O$
So, 3g of ${{H}_{2}}$ produces $=\dfrac{36}{4}\times 3=27g$
Therefore, 27g of water is produced when 3g of Hydrogen is used.

-(iii) As 24g of oxygen is utilized in the reaction out of 29g of given oxygen,
So, the amount of oxygen which is unreacted = 29 – 24 = 5g
Therefore 5gm of Oxygen is unreacted in the process.

Note: There are two methods for finding and calculating the limiting agent and its amount in a chemical reaction.
(i) The first method is by the comparison of reactant amounts. This method is more useful when there are only two reactants present in the chemical reactant. In this method, one reactant (let us say A) is chosen and then the balanced chemical equation is used to determine the amount of the other reactant (let us say B) necessary to react with the first reactant (A). If the amount of B present is more than the required amount, then the reactant B is said to be in excess and thus A will be limiting reagent.
(ii) The second method is by comparison of product amounts which can be formed from each reactant. In this method, the chemical equation is used to calculate the amount of product formed from the amount of reactant present. The limiting reagent is the one that will form the smallest amount of the product. This method can be extended to find the product formed by any number of the reactants.