Question

Phosphoric acid $\left({\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\right)$ prepared in a two step process.
(1) ${\text{P}}_{\text{4}}{\text{+ 5O}}_2\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$
(2) ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}+{\text{6H}}_{\text{2}}\text{O}\to {\text{4H}}_{\text{3}}{\text{PO}}_{\text{4}}$
We allow 62 g of phosphorus to react with excess oxygen which form ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ which gives 85% yield. If in the step (2) reaction, 90% yield of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is obtained then mass of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$produced is:

• A. 37.485 g
• B. 149.94 g
• C. 125.47 g
• D. 564.48 g

Answer 1

The correct option is B 149.94 g

Moles of $P_4=\frac{62}{124}=0.5$

$P_4+5O_2⟶P_4{O}_{10}$

$0.5$ $moles$

Moles of $P_4{O}_{10}$ formed $=0.85$ (yield) $\times 0.5=0.425$ $moles$

$P_4{O}_{10}+6H_{2}O⟶H_{3}P{O}_4$

$0.425$ $moles$

Moles of $H_{3}P{O}_4$ formed $=0.425\times 0.9$ (yield) $=0.3825$ $moles$

Mass of $H_{3}P{O}_4$ formed $=98\times 0.3825=37.485$ $g$

Hence, the correct option is $\left(A\right)$.