Phosphorus may be determined analytically by precipitating it as MgNH4PO4 followed by ignition to Mg2P2O7. What mass of Mg2P2O7 will be obtained from a 9.346 g sample of phosphate rock which contains 71.24% Ca3(PO4)2?

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Chapter4: Stoichiometry
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Phosphorus may be determined analytically by precipitating it as MgNH4PO4 followed by ignition to Mg2P2O7. What mass of Mg2P2O7 will be obtained from a 9.346 g sample of phosphate rock which contains 71.24% Ca3(PO4)2?

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Step 1

Given data:

Mass of sample of phosphate rock = 9.346 gms

It contains 71.24% of Ca3(PO4)2

So, mass of Ca3(PO4)2 in sample rock = 71.24100×9.346=6.66 gms

Molar mass of Ca3(PO4)2 = 310.17 gms/mol

Mass of Phosphorous in Ca3(PO4)2 = 2 x 30.97 = 61.94 gms

So, mass of phosphorous in sample = 6.66×61.94310.17=1.32 gms

Now, Number of moles of P=Mass of PMolecular Mass of P=1.3230.97=0.043 moles

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