Question

How many gram of $H_{3}P{O}_4$ would be needed to neutralise 100 gm of $Mg(OH{)}_2$?

• A. 66.7 gm
• B. 252 gm
• C. 112.6 gm
• D. 168 gm

Answer 1

The correct option is C 112.6 gm

$2H_{3}P{O}_4+3Mg(OH{)}_2\to M{g}_3(P{O}_4{)}_2+6H_{2}O$

Moles of $Mg(OH{)}_2=\frac{100}{58}$

Moles of $H_{3}P{O}_4$ needed $=\frac{100}{58}\times \frac{2}{3}$

Weight of $H_{3}P{O}_4$ needed $=\frac{100}{58}\times \frac{2}{3}\times 98=112.6gm$