Question

How many grams of phosphoric acid would be needed to neutralise $100$ g of magnesium hydroxide? (The molecular weights of $H_{3}P{O}_4=98$ g/mol and $Mg(OH{)}_2=58.3$ g/mol )

• A. $66.7$ g
• B. $252$ g
• C. $112$ g
• D. $168$ g

Answer 1

Answer: (C)

$112$ g

The number of moles of magnesium hydroxide are $\frac{100}{58.3}=1.72$.
The number of moles of phosphoric acid required for neutralization are $\frac{2}{3}\times 1.72=1.14$ mol.
The mass of phosphoric acid required is $1.14\times 98=112g$.