the balanced equation for the above reaction is as follows;
3Ca(OH)₂ + 2H₃PO₄ ---> Ca₃(PO₄)₂ + 6H₂O
stoichiometry of Ca(OH)₂ to Ca₃(PO₄)₂ is 3:1
Since Ca(OH)₂ is the limiting reactant, amount of Ca₃(PO₄)₂ formed depends on the amount of Ca(OH)₂ present
Number of Ca(OH)₂ reacted - 21.5 g/74 g/mol = 0.29 mol
Therefore number of Ca₃(PO₄)₂ moles formed - 0.29 mol/3 = 0.0967 mol
Mass of Ca₃(PO₄)₂ formed - 0.0967 mol x 310 g/mol = 30.0 g