Question

10 g of $NaOH$ required certain amount of $H_{2}S{O}_4$ for complete neutralization. Calculate the amount of $H_{2}S{O}_4$ required.

• A. 10 g
• B. 12.25 g
• C. 24.5 g
• D. 49 g

Answer 1

The correct option is B 12.25 g

10 g of $NaOH$ (molar mass 40 g/mol) $=\frac{10g}{40g/mol}=0.25mole$

$2NaOH+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2H_{2}O$

0.25 moles of $NaOH$ will react with $\frac{0.25}{2}=0.125$ moles of $H_{2}S{O}_4$.

Molar mass of $H_{2}S{O}_4=$ 98 g/mol.

Mass of $H_{2}S{O}_4=0.125mol\times 98g/mol=12.25$ g.

Hence, the correct option is B.