Question

Phosphoric acid is prepared in a two-step process. Find the mass of $H_{3}P{O}_4$ formed when 124 g of $P_4$ reacts with an excess of oxygen and then treated with sufficient water.
$P_4+O_2⟶P_4{O}_{10}$
$P_4{O}_{10}+H_{2}O⟶H_{3}P{O}_4$
(Molar mass of P = 31 g/mol)

• A. 150 g
• B. 288 g
• C. 560 g
• D. 392 g

Answer 1

The correct option is D 392 g

Let the weight of acid produced be $Xg$.
Since the atoms of $P$ are conserved,
Applying POAC on phosphorus,
$4\times \text{moles of}{P}_4=1\times \text{moles of}{H}_{3}P{O}_4$

$\Rightarrow 4\times \frac{\text{weight of}{P}_4}{\text{molar mass of}{P}_4}=1\times \frac{\text{weight of acid}}{\text{molar mass of acid}}$

$\Rightarrow 4\times \frac{124}{124}=1\times \frac{X}{98}$

On solving, $X=392g$
Therefore,the mass of phosphoric acid produced is 392 g.