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Calcuim phosphide (Ca3P2) formed by reac...

Calcuim phosphide `(Ca_3P_2)` formed by reacting calcuim orthophosphate `(Ca_3(PO_4)_2)` with magnesium was hydrolsed by water. The evolved phosphine `(PH_3)` was burnt in air to yield phosphorus pentoxide `(P_2O_5)`. How many grams of magnesium metaphosphate would obtained? If 19.2 g of magnesium were used for reducing calcium phosphate (At. wt Mg =24, P=31)
`Ca_3(PO_4)_2+Mg toCa_3P_2+MgO`
`Ca_3P_2+H_2OtoCa(OH)_2+PH_3`
`PH_3+O_2toP_2O_5+H_2O`
`MgO+P_2O_5tounderset("magnesium metaphosphate")(Mg(PO_3)_2)`

Text Solution

Verified by Experts

The correct Answer is:
18 gram

Balance chemical equations are :
`Ca_3(PO_4)_2+8Mgt OCa_3P_2+8MgO`
`Ca_3P_2+6H_2Oto3Ca(OH)_2+2PH_2`
`2PH_3+4O_2toP_2O_5+3H_2O`
`MgO+P_2O_5toMg(PO_3)_2`
moles of magnesium used =0.8 moles
moles of MgO formed =0.8 moles
moles of `Ca_3P_2` formed 0.1 moles
moles of `PH_3` formed=0.2 moles
moles of `P_2O_5` formed =0.1 moles (limiting reagent)
moles of `Mg(PO_3)_2=0.1` moles
mass of `Mg(PO_3)_2=18.2` gram
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Calcium phosphide Ca_(3)P_(2) formed by reacting magnesium with excess calcium orthophosphate Ca_(s)(PO_(4))_(2) was hydrolysed by excess water. The evolved. Phosphine PH_(5) was burnt in air to yield phosphrous pentoxide (P_(2)O_(6)) . How many gram of magnesium metaphosphate would be obtain if 192 gram Mg were used (Atomic weight of Mg=24, P=31) Ca_(3)(PO_(4))_(2) + Mg to Ca_(3)P_(2) +_ MgO Ca_(3)P_(2)+H_(2)O to Ca(OH)_(2) + PH_(3) PH_(3) + O_(2) to P_(2)O_(5) + H_(2)O MgO + P_(2)O_(5) to Mg(PO_(3))_(2) " " magnesium metaphosphate.

Ca_(3)(PO_(4))_(2) is :

Knowledge Check

  • How many moles of magnesium phosphate Mg_3 (PO_4)_2 will contain 0.25 mole of oxygen atoms ?

    A
    (a) `0.02`
    B
    (b) `3.125 xx 10^(-2)`
    C
    (c) `1.25 xx 10^(-2)`
    D
    (d) `2.5 xx 10^(-2)`
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    Ca_(3)(PO_(4))_(2) is :

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    Phosphorus pentoxide NH_(3)+P_(2)O_(5) + H_(2)O to______

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