# A 6.681-g sample containing magnesium and sodium chlorides was dissolved in sufficient water to give 500.0 mL?

of solution. Analysis for the chloride content of 50.00-mL aliquot resulted in the formation of 0.5923 g of AgCl. The Mg in the second 50.00-mL aliquot was precipitated as MgNH(subscript)4PO(sub)4; upon ignition 0.1796 g of Mg(sub)2P(sub)2O(sub)7 was found. Calculate percentage of MgCl(sub)2 and of NaCl in the sample. (MgCl(sub)2 = 95.22 g/mol, NaCl = 58.44 g/mol, Mg(sub)2P(sub)2O(sub)7 = 222.57 g/mol, AgCl = 143.32 g/mol)

this is a problem on gravimetric calculations. please help.Ü

Answer:
Moles Mg2P2O7 = 0.1796 / 222.57 = 0.0008069
Moles MgNH4PO4 = 2 x 0.0008069 = 0.001614 = moles Mg2+ in 50.00 mL
In 500 mL there are 0.01614 moles
0.01614 mol x 95.22 g/mol = 1.537 g MgCl2
1.537 : 6.681 = x : 100
x = 23.00 % of MgCl2 in the sample

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