Calcium phosphide (Ca 3 P 2 ) formed by reacting calcium orthophosphate (Ca 3 (PO 4 ) 2 ) with magnesium was hydrolyzed by water. The evolved phosphine (PH 3 ) was burnt in air to yield phosphorus pentoxide (P 2 O 5 ). How many grams of magnesium metaphosphate would be obtained, if 19.2 g of magnesium were used for reducing calcium phosphate.
Ca 3 (PO 4 ) 2 + Mg 
Ca 3 P 2 + MgO
Ca 3 P 2 + H 2 O 
Ca(OH) 2 + PH 3
PH 3 + O 2 
P 2 O 5 + H 2 O
MgO + P 2 O 5 
Mg(PO 3 ) 2
magnesium metaphosphate
Text Solution
Verified by ExpertsD
Balance chemical equations are :
Ca 3 (PO 4 ) 2 + 8Mg 
Ca 3 P 2 + 8MgO
Ca 3 P 2 + 6H 2 O 
3Ca(OH) 2 + 2PH 3
2PH 3 + 4O 2 
P 2 O 5 + 3H 2 O
MgO + P 2 O 5 
Mg(PO 3 ) 2
moles of magnesium used = 0.8 moles
moles of MgO formed = 0.8 moles
moles of Ca 3 P 2 formed 0.1 moles
moles of PH 3 formed = 0.2 moles
moles of P 2 O 5 formed = 0.1 mole (limiting reagent)
moles of Mg(PO 3 ) 2 = 0.1 moles
mass of Mg(PO 3 ) 2 = 18.2 gram
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