Start with a balanced equation, which will give you the mole ratios for the reactants and products.
"2Na"_3"PO"_4("aq")2Na3PO4(aq) + "3Ca(OH")_2("aq")3Ca(OH)2(aq) rarr→ "Ca"_3("PO"_4)_2("s")Ca3(PO4)2(s) + "6NaOH"("aq")6NaOH(aq)
Assumption: "Ca(OH)"_2Ca(OH)2 is in excess so that "Na"_3"PO"_4Na3PO4 is the limiting reactant.
First, convert the mass of sodium phosphate to moles. The molar mass of "Na"_3"PO"_4Na3PO4 is "163.94g/mol"163.94g/mol.
"5.640g Na"_3"PO"_45.640g Na3PO4 x "1 mol"/"163.94g"1 mol163.94g = "0.0344028mol Na"_3"PO"_40.0344028mol Na3PO4
Second, calculate the moles of "Ca"_3("PO"_4)_2Ca3(PO4)2 produced by multiplying the moles of "Na"_3"PO"_4Na3PO4 times the mole ratio of "Na"_3"PO"_4Na3PO4 to "Ca"_3("PO"_4)_2Ca3(PO4)2 from the balanced equation, in which "Ca"_3("PO"_4)_2Ca3(PO4)2 is on top.
"0.0344028mol Na"_3"PO"_40.0344028mol Na3PO4 x "1 mol Ca"_3("PO"_4)_2"/2 mol Na"_3"PO"_41 mol Ca3(PO4)2/2 mol Na3PO4 = "0.172014mol Ca"_3("PO"_4)_20.172014mol Ca3(PO4)2
Third, convert moles of "Ca"_3("PO"_4)_2Ca3(PO4)2 to mass . The molar mass of "Ca"_3("PO"_4)_2Ca3(PO4)2 is "310.18g/mol"310.18g/mol.
"0.172014mol Ca"_3("PO"_4)_20.172014mol Ca3(PO4)2 x "310.18g"/"1 mol"310.18g1 mol = "53.36g Ca"_3("PO"_4)_253.36g Ca3(PO4)2 (rounded to four significant figures due to four significant figures in "5.640g"5.640g)
Answer:
Assuming that "Ca(OH)"_2Ca(OH)2 was in excess, "5.640g Na"_3"PO"_45.640g Na3PO4 will produce "53.36g Ca"_3("PO"_4)_253.36g Ca3(PO4)2 in the reaction described above.