Question

The mass of $70\%$ $H_{2}S{O}_4$ required for the neutralization of $1$ mole of $NaOH$ is :

• A. $70$ g
• B. $35$ g
• C. $30$ g
• D. $95$ g

Answer 1

Answer: (A)

$70$ g

$NaOH+H_{2}S{O}_4⟶N{a}_{2}S{O}_4+2H_{2}O$
$2$ mol $1$ mol
$1$ mol $0.5$ mol $=49$ g $H_{2}S{O}_4$ (pure)
Thus, $70\%$ $H_{2}S{O}_4$ required $=49\times \frac{100}{70}=70$ g