balanced chemical equation is
3CaCl2 +2Na3PO4-->6NaCl +Ca3(PO4)2
moles of CaCl2 =89.3g/[ (35.5x2) +40]=0.805moles
from the equation above the ratio of CaCl2 to Ca3(PO4)2 is 3:1 therefore the moles of Ca3(PO4)2 is 0.809/3=0.268moles
mass is therefore 0.268 x310.18(R.F.M of Ca3(PO4)2 ) =83.23grams