A sample containing 0.150 g of magnesium (as $\mathrm{Mg}{}^{2+}$) is precipitated as $\mathrm{Mg}(\mathrm{NH}{}_4)\mathrm{PO}{}_4\cdot 6\mathrm{H}{}_2\mathrm{O}$ and ignited strongly: $2\mathrm{Mg}(\mathrm{NH}{}_4)\mathrm{PO}{}_4\to \mathrm{Mg}{}_2\mathrm{P}{}_2\mathrm{O}{}_7+2\mathrm{NH}{}_3+\mathrm{H}{}_2\mathrm{O}$. Atomic masses: Mg 24, P 31, O 16. The mass of ignited residue is

Moles of Mg = 0.150/24 = 6.25 mmol. Two Mg give one $\mathrm{Mg}{}_2\mathrm{P}{}_2\mathrm{O}{}_7$ (molar mass 2(24)+2(31)+7(16) = 222 g/mol), so moles of residue = 3.125 mmol. Mass = 0.00312 x 222 = 0.694 g. The answer 1.39 g wrongly assumes one pyrophosphate per Mg, and 0.250 g would be $\mathrm{MgO}$.