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.
