Question

It has been claimed that $N{H}_{4}N{O}_3$ fertiliser can be rendered unexplodable by a process that involves additives such as diammonium hydrogen phosphate $(NH{)}_4{)}_{2}HP{O}_4$. Analysis of such a "densensitised" sample of $N{H}_{4}N{O}_3$ showed the mass per cent nitrogen to be 33.81%. % of $N{H}_{4}N{O}_3$ in this fertilizer is?

• A. $N{H}_{4}N{O}_3=88\%$
• B. $N{H}_{4}N{O}_3=95\%$
• C. $N{H}_{4}N{O}_3=81\%$
• D. $N{H}_{4}N{O}_3=91\%$

Answer 1

The correct option is D $N{H}_{4}N{O}_3=91\%$

let
$(N{H}_4{)}_{2}HP{O}_4=x$
$N{H}_{4}N{O}_3=(100-x)$
Nitrogen in $(N{H}_4{)}_{2}HP{O}_4=\frac{28x}{132}$
Nitrogen in $N{H}_{4}N{O}_3=\frac{28(100-x)}{80}$
$\therefore \frac{28x}{132}+\frac{28(100-x)}{80}=33.81$
$\frac{x}{132}+\frac{100-x}{80}=1.2075$
$x=8.63$
$\therefore (N{H}_4{)}_{2}HP{O}_4=8.63$
$\therefore N{H}_{4}N{O}_3=91.37$