Question

If 1.11 g $M{g}_2{P}_2{O}_7$ is formed at the end of the reaction, then what weight of $Na{H}_{2}P{O}_4$ was present originally given that no other products contained phosphorus?
$\text{(Molar mass}M{g}_2{P}_2{O}_7=222{\text{g mol}}^{-1},Na{H}_{2}P{O}_4=120{\text{g mol}}^{-1})$
$Na{H}_{2}P{O}_4+M{g}^{2+}+N{H}_4^{+}\to Mg\left(N{H}_4\right)P{O}_4.6H_{2}O\to M{g}_2{P}_2{O}_7$

Answer 1

$Na{H}_{2}P{O}_4+M{g}^{2+}+N{H}_4^{+}\to Mg\left(N{H}_4\right)P{O}_4.6H_{2}O\to M{g}_2{P}_2{O}_7$
Applying POAC on P,
Moles of P in $Na{H}_{2}P{O}_4$= moles of P in $M{g}_2{P}_2{O}_7$
$1\times \text{number of mol of}Na{H}_{2}P{O}_4=2\times \text{number of mol of}M{g}_2{P}_2{O}_7$
$\frac{\text{Weight of}Na{H}_{2}P{O}_4}{\text{molar mass}}=2\times \frac{\text{Weight of}M{g}_2{P}_2{O}_7}{\text{molar mass}}\frac{\text{Weight of}Na{H}_{2}P{O}_4}{120}=2\times \frac{1.11}{222}$
Weight of $Na{H}_{2}P{O}_4$ =1.2 g
Thus, the weight of $Na{H}_{2}P{O}_4$ present originally was 1.2 g