Question

How many grams of nickel (II) phosphate (366 g/mol) will precipitate if excess sodium phosphate is added to 1.20 L of 0.684 M nickel acetate?

(a) 100 g

(b) 200 g

(c) 300 g

(d) 600 g

(e) 167 g

Answer 1

3Ni(CH3COO)2 (aq) + 2Na3PO4(aq) --------------------> Ni3(PO4)2(s) + 6CH3COONa(aq)

no of moles of Ni(CH3COO)2 = molarity * volume in L

                                                  = 0.684*1.2   = 0.8208 moles

3moles of Ni(CH3COO)2 react with Na3PO4 to gives 1 mole of Ni3(PO4)2

0.8208 moles of Ni(CH3COO)2 react with Na3PO4 to gives    = 1*0.8208/3   = 0.2736 moles of Ni3(PO4)2

mass of Ni3(PO4)2   = no of moles * gram molar mass

                                    = 0.2736*366    = 100g

a) 100 g >>>>answer