Question

A certain sample of phosphate rock contains $26.26\%$ $P_2{O}_5$. A $0.5428$ g sample is analysed by precipitating $MgN{H}_{4}P{O}_4.6H_{2}O$ and igniting the precipitate to $M{g}_2{P}_2{O}_7$. Thus, $M{g}_2{P}_2{O}_7$ obtained is :

• A. $0.8486$ g
• B. $0.1424$ g
• C. $0.3648$ g
• D. $0.2228$ g

Answer 1

The correct option is D $0.2228$ g

$P_2{O}_5⟶M{g}_2{P}_2{O}_7$
$1$ mol $1$ mol
$142$ g $222$ g
$P_2{O}_5$ in the mineral $=0.5428\times \frac{26.26}{100}$ $=0.1425$ g $=1\times {10}^{-3}$ mol
Thus, $M{g}_2{P}_2{O}_7$ obtained $=1\times {10}^{-3}$ mol $=0.2228g$