Start with a balanced equation, which will give you the mole ratios for the reactants and products.
2Na3PO4(aq) + 3Ca(OH)2(aq) → Ca3(PO4)2(s) + 6NaOH(aq)
Assumption: Ca(OH)2 is in excess so that Na3PO4 is the limiting reactant.
First, convert the mass of sodium phosphate to moles. The molar mass of Na3PO4 is 163.94g/mol.
5.640g Na3PO4 x 1 mol163.94g = 0.0344028mol Na3PO4
Second, calculate the moles of Ca3(PO4)2 produced by multiplying the moles of Na3PO4 times the mole ratio of Na3PO4 to Ca3(PO4)2 from the balanced equation, in which Ca3(PO4)2 is on top.
0.0344028mol Na3PO4 x 1 mol Ca3(PO4)2/2 mol Na3PO4 = 0.172014mol Ca3(PO4)2
Third, convert moles of Ca3(PO4)2 to mass . The molar mass of Ca3(PO4)2 is 310.18g/mol.
0.172014mol Ca3(PO4)2 x 310.18g1 mol = 53.36g Ca3(PO4)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 Na3PO4 will produce 53.36g Ca3(PO4)2 in the reaction described above.
