Question

How many moles of $KOH$ are required to neutralize $196$ grams of sulphuric acid? ( Given $H_{2}S{O}_4=98$ amu$)$?

• A. $1.0$
• B. $1.5$
• C. $2.0$
• D. $4.0$

Answer 1

The correct option is D $4.0$

The atomic mass of $H_{2}S{O}_4=98$ g/mol

The number of moles of $H_{2}S{O}_4$ in $196$ grams $=\frac{196g}{98g/mol}=2$ moles

1 mole of $H_{2}S{O}_4$ will be neutralized by $2$ moles of $KOH$.

$H_{2}S{O}_4+2KOH\to K_{2}S{O}_4+2H_{2}O$

Hence, 2 moles of $H_{2}S{O}_4$ will be neutralized by $4$ moles of $KOH$.