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