Question

Phosphoric acid ($H_{3}P{O}_4$) is prepared in a two-step process.
$P_4+5O_2\to P_4{O}_{10}$
$P_4{O}_{10}+6H_{2}O\to 4H_{3}P{O}_4$
186 g of phosphorus is made to react with excess oxygen which forms $P_4{O}_{10}$ that has a 60 % yield. In the second reaction, 80% yield of $H_{3}P{O}_4$ is obtained. The mass of $H_{3}P{O}_4$ produced is:
(Molar mass of phosphorous = 31 g/mol)

• A. 282.24 g
• B. 149.95 g
• C. 125.47 g
• D. 564.48 g

Answer 1

The correct option is A 282.24 g

Moles of $P_4=\frac{186}{124}=1.5mol$

$P_4+5O_2\to P_4{O}_{10}$
1.5
Yield of the reaction is given as 60%.
Moles of $P_4{O}_{10}$ formed $=0.60\times 1.5=0.90mol$

$P_4{O}_{10}+6H_{2}O\to 4H_{3}P{O}_4$
0.90
1 mole of $P_4{O}_{10}$ produces 4 moles of $H_{3}P{O}_4$
Yield of the reaction is given as 80%
Moles of $H_{3}P{O}_4$ formed $=0.90\times 0.80\times 4=2.88mol$
Mass of $H_{3}P{O}_4$ formed $=98\times 2.88=282.24g$