Question

write the formula and calculate the mass of the precipitate formed when excess potassium phosphate is mixed with 55.0 ml of 0.650 M of magnesium nitrate

Answer 1

Balanced equation :-

2 K₃PO_4(aq)+3 Mg(NO₃)_2(aq) ------> Mg₃(PO₄)_2(s)+

6 KNO3_(aq)

Formula for precipitate = Mg₃(PO₄)₂

Concentration of Mg(NO3)2 = 0.650 M = 0.650 mol/L

Volume of Mg(NO3)2 solution = 55.0 ml = 55.0 L / 1000 = 0.055 L

Number of moles of Mg(NO3)2 = 0.055 L * 0.650 mol/L = 0.03575 mol

From reaction, 3.0 moles of Mg(NO3)2 produces 1.0 mol of Mg3(PO4)2 so 0.03575 mol of Mg(NO3)2 will produce 1.0 * 0.03575 mol / 3 = 0.012 mol of Mg3(PO4)2.

Molar mass of Mg3(PO4)2 = 262.86 g/mol

Mass of precipitate formed = 262.86 g/mol * 0.012 mol

= 3.15 g