Question

Sulphuric acid reacts with sodium hydroxide as follows
${H_2}S{O_4} + \,2NaOH \to N{a_2}S{O_4} + 2{H_2}O$
When $1\,\,L$ of $0.1\,\,M$ sulphuric acid solution is allowed to react with $1\,\,L$of \[0.1\,\,M\] sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is?
A. $0.1\,\,mol{L^{ - 1}}$
B. $7.10\,g$
C. $0.025\,\,mol{L^{ - 1}}$
D. $3.55\,g$

Answer 1

Hint:
Here, the concept of limiting reagent is used. Limiting reagents is that reagent in the reaction which consumes first and it will present in a lesser amount.

Complete step by step solution
The limiting reagent provides the actual yield of the product that is formed from the reaction. To find the limiting reagent, first calculate the number of moles of each reagent. The reagent which contains a lower number of moles is the limiting reagent.
Molarity is calculated as the number of moles per liter. It is denoted by $M$ .$1\,M$ means one mole of a substance dissolved in one litre of solution. If molarity is $5\,M$, then it means $5\,moles$ of substance dissolved in 1 litre.
Given that;
The number of moles of ${H_2}S{O_4}$ is $0.1\,\,moles$
The number of moles of \[NaOH\] is \[0.1\,\,moles\]
The reaction takes place as;
${H_2}S{O_4} + \,2NaOH \to N{a_2}S{O_4} + 2{H_2}O$

From the stoichiometric coefficient’s chemical reaction, it can be seen one mole of ${H_2}S{O_4}$ and two moles of \[NaOH\] reacts. So, they react in the ratio of 1:2.
So according to the ratio, $0.1\,\,moles$ of ${H_2}S{O_4}$ reacts with $0.1 \times 2\,moles = 0.2\,moles$ of \[NaOH\] which is not present so it is the limiting reagent.
But in the reaction $0.1\,\,moles$ of \[NaOH\] will react with \[0.05\,moles\] of ${H_2}S{O_4}$.
Since \[NaOH\] is limiting reagent in our reaction so the product will be formed according to\[NaOH\].
So,$N{a_2}S{O_4}$ formed is equal to \[0.05\,moles\].
Molar mass of $N{a_2}S{O_4}$is calculated as shown below.
$
N{a_2}S{O_4} = 2 \times 23 + 32 \times 1 + 16 \times 4\\
 = 142\,g\;mo{l^{ - 1}}
$
Mass of $N{a_2}S{O_4}$ is calculated as shown below.
$
{\rm{Mass}} = 0.05 \times 142\,g\\
 = 7.1\;g
$

Hence, option B is correct.

Note:
The reaction involved in this question is an acid-base neutralisation reaction where both the acid and the base are strong.If the acid is the limiting reagent then the resulting solution is basic in nature and vice-versa.