Question

The maximum amount of $BaS{O_4}$ that can be obtained on mixing $0.5$ mole of $BaC{l_2}$ with one mole of ${H_2}S{O_4}$ is:
A. $0.5$ mole
B. $0.1$ mole
C. $0.15$ mole
D. $0.2$ mole

Answer 1

Hint: In the reaction, reactants are on the left side of the arrow and products are on the right side of the arrow. If we know the concentration of reactants then after writing the balanced equation we can compare how much moles of reactant will generate how much amount of product.

Complete step by step answer:
First of all we will read about reactants and products.
Reactants: The compounds which react with each other to form the desired compounds, are known as reactants. They are present on the left side of the arrow in the reaction.
Products: The compounds which are formed after the reaction of reactants, are known as products. They are present on the right side of the arrow in the reaction.
Now, we will see what is a balanced reaction?
Balanced reaction: The reaction in which the number of atoms of each element in both the sides i.e. on the reactant side as well as on the product side are same, is known as balanced reaction.
Here we are given with the reaction of barium chloride with sulphuric acid. The reaction between these two reactants are as follows:
${\text{BaC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{BaS}}{{\text{O}}_{\text{4}}}{\text{ + 2HCl}}$. This is the balanced reaction between barium chloride and sulphuric acid as all atoms on both sides are the same in number.
From the balanced reaction of barium chloride and sulphuric acid we can see that one mole of barium chloride will react with one mole of sulphuric acid to form one mole of barium sulphate. So, half mole of barium chloride will react with half mole of sulphuric acid to form half mole of barium sulphate.
Here in the question it is given that half mole of barium chloride and mole of sulphuric is present in the starting of the reaction. Then half mole of barium chloride will react with half mole of sulphuric acid to form half mole of barium sulphate. And half mole of sulphuric will remain unreacted.
So, half i.e. $0.5$ mole of barium sulphate will be formed on the reaction of half mole of barium chloride with one mole of sulphuric acid.

So, the correct answer is Option A .

Note:
The extra amount will remain unused after the reaction. For example: In the above reaction the half mole of sulphuric acid will remain unused after the reaction. It will be there in the container in which the reaction is going on along with the product.