Question

$PC{l}_5+4H_{2}O\to H_{3}P{O}_4+5HCl$

$n{H}_{3}P{O}_4\underset{upto{316}^{\circ }C}{\overset{heat}{\to }}(HP{O}_3{)}_n+n{H}_{2}O$

The mass of PCl${}_5$ required to prepare 320 g of cyclic metaphosphoric acid ${\left(HP{O}_3\right)}_n$ is:

• A. 208.5 g
• B. 417 g
• C. 1042.5 g
• D. 834 g

Answer 1

The correct option is D 834 g

One mole of phosphorus pentachloride will produce one mole of metaphosphoric acid.

The molar masses of phosphorus pentachloride and metaphosphoric acid are $208.24g/mol$ and $80g/mol$ respectively.

320 g of metaphosphoric acid will be obtained from $\frac{208.24}{80}\times 320=834g$ of phosphorus pentachloride respectively.

Hence, option D is correct.