Question

In diammonium hydrogen phosphate, $(N{H}_4{)}_{2}HP{O}_4$, percentage of :

• A. $P_2{O}_5$ is $53.78$%
• B. $N{H}_3$ is $25.76$%
• C. $P$ is maximum
• D. $N$ is maximum

Answer 1

Answer: (A), (B)

$P_2{O}_5$ is $53.78$%
$N{H}_3$ is $25.76$%

$1$ mole of $(N{H}_4{)}_{2}HP{O}_4$ has $0.5$ mole $P_2{O}_5$ and $2$ of $N{H}_3$.
Thus, $1$ mole ($132$ g) $(N{H}_4{)}_{2}HP{O}_4$ has $0.5$ mole ($71$ g) $P_2{O}_5$
$\%$ of $P_2{O}_5=\frac{71}{132}\times 100=53.78$%
Also, $132$ g $(N{H}_4{)}_{2}HP{O}_4$ has $34$ g $N{H}_3$.
Thus, $\%$ of $N{H}_3=\frac{34}{132}\times 100=25.76$%
Also, $132$ g $=28$ g $N$, $9$ g $H$ and $64$ g $O$
Thus, $\%$ of $O$ is maximum.