Question

How many grams of $H_{3}P{O}_4$ is required to completely neutralise 120 g of $NaOH$?

• A. $49$
• B. $98$
• C. $196$
• D. $9.8$

Answer 1

The correct option is B $98$

The balanced equation for the reaction is:

$H_{3}P{O}_4+3NaOH\to N{a}_{3}P{O}_4+3H_{2}O$

$\therefore$ Amount of $H_{3}P{O}_4$ required to neutralise 120g of NaOH = Molecular wt. of $H_{3}P{O}_4$ = ($1\times 3+31+16\times 4$) = $98g$