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Excess of calcium orthophosphate is reacted with magnesium to form calcium phosphide $({Ca}_{3} P_{2})$ along with magnesium oxide. Calcium phosphide on reacting with excess of water liberate phosphine gas $({PH}_{3})$ along with calcium hydroxide. Phosphine is burnt in excess of oxygen to form $P_{2} O_{5}$ along with water. Oxides of magnesium and phosphorous react to give magnesium metaphosphate. Calculate grams of magnesium metaphosphate obtained if 1.92 gm of magnesium is taken.
[Round off your answer to nearest integer]

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answer is 2.

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Detailed Solution

$8 Mg + {Ca}_{3} {({PO}_{4})}_{2} \to {Ca}_{3} P_{2} + 8 MgO$

$\frac{1.92}{24} \frac{1.92}{24} \times \frac{1}{8} \frac{1.92}{24}$

${Ca}_{3} P_{2} + 6 H_{2} O \to 2 {PH}_{3} + 3 Ca (OH)_{2}$

$2 \times \frac{1.92}{24} \times \frac{1}{8} \frac{1.92}{24} \times \frac{1}{8}$

$2 {PH}_{3} + 4 O_{2} \to P_{2} O_{5} + 3 H_{2} O$

$2 \times \frac{1.92}{24} \times \frac{1}{8} \frac{1.92}{24} \times \frac{1}{8}$

$MgO + P_{2} O_{5} \to Mg {({PO}_{3})}_{2}$

$\frac{1.92}{24} \frac{1.92}{24} \times \frac{1}{8} \frac{1.92}{24} \times \frac{1}{8} \times 182 g$

(LR) = 1.82 g

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