Question

The mass of 70% ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ required for neutralization of one mole of NaOH.
(A) 49g
(B) 98g
(C) 70g
(D) $34.3$ g

Answer 1

Write the neutralization reaction of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ and NaOH. Find the ratio of no. of moles of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ and NaOH required for the reaction. Then find the no. of moles of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ corresponding to one mole of NaOH. According to the no. of moles of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ being used find its mass and then obtain the mass of 70% ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ .

Complete Step By Step Solution
The neutralization reaction is given as- $2\text{NaOH}+{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\to \text{N}{{\text{a}}_{2}}\text{S}{{\text{O}}_{4}}+2{{\text{H}}_{2}}\text{O}$ , the reaction of sodium hydroxide and sulfuric acid gives sodium sulfate and water. As we can see from the reaction that 2 moles of NaOH and 1 mole of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ are used for the neutralization process. So if 2 moles of NaOH corresponds to 1 mole of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ then I mole of NaOH will correspond to $0\cdot 5$ mole of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ .
The molar mass or mass for 1 mole of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ -
\implies\ 2\times 1+32+16\times 4 \\\ \implies2+32+64 \\\ \implies 98g \\\
So, mass of 0.5 mole of ${{H}_{2}}S{{O}_{4}}$ = 49g.
Therefore mass of 70 ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$
\implies\dfrac{49\times 100}{70} \\\ \implies70g \\\
\implies\dfrac{49\times 100}{70} \\\ \implies\dfrac{490}{7} \\\ \implies70g \\\

So, the correct choice is (C).

Additional Information
Neutralization reaction is a chemical reaction in which acid and base reacts quantitatively with each other to form salt and water as end products. Water is formed in the neutralization reaction as hydrogen ions from acid and hydroxyl ions from base combine together. In the above given neutralization reaction, strong sulfuric acid and strong base sodium hydroxide reacts together to form sodium sulfate as salt and the water as products. There are many applications of neutralization reactions which includes-
Titration Method- It is a chemical analysis in which the concentration of an unknown solution known as analyte is found by adding the solution with known concentration which is known as titrate. The indicator is used at the equivalence point which indicates the neutralization reaction that undergoes between the added titrant and analyte solution.
It is used in the nanomaterial synthesis where heat of neutralization reaction is used for the chemical reduction of metal precursors (contains metal ions).
Digestive System of the human body also undergoes a neutralization reaction in the stomach to neutralize the food.
The acidity of the soil is also controlled by adding suitable bases that undergo neutralization reactions in order to control the pH of the soil to obtain the optimal conditions required for plant growth.

Note
Check the correct ratio of the no. of moles of acid and base required for neutralization reaction. Balance the chemical reaction in order to obtain the correct result. The 70% mass of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ equals the mass of $0\cdot 5$ mole of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ by the total mass required for neutralization i.e.
70%=$\dfrac{49}{\text{mass required}}$
$\dfrac{70}{100}$=$\dfrac{49}{\text{mass required}}$
$\text{mass required}$=$\dfrac{49\times 100}{70}$