Question

How many grams of phosphoric acid would be needed to neutralise $100$ g of magnesium hydroxide? (The molecular weights of $H_{2}P{O}_4=98$ and $Mg(OH{)}_2=58.3$ g/mol respectively)

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

Answer 1

The correct option is D $168$ g

As the neutralizing reaction between $H_{2}P{O}_4$ and $Mg(OH{)}_2$ is as follows:
$H_{2}P{O}_4+Mg(OH{)}_2\to MgP{O}_4+2H_{2}O$
this reaction shows that 1 mol of $H_{2}P{O}_4$ neutralises 1 mol of $Mg(OH{)}_2$
now to determine the moles of these from given weight of them
moles of $Mg(OH{)}_2$ = Weight/ molar mass = $100/58.3=1.72$ moles
moles of $H_{2}P{O}_4$ = Weight/ molar mass = x/98 moles ( let wt = x)
so as moles of $H_{2}P{O}_4$ and $Mg(OH{)}_2$ must be same., as per balanced equation.
so $x/98=1.72$
so, $x=98\ast 1.72=168.56$ hence nearby 168 g of $H_{2}P{O}_4$.