The grams  of calcium phosphate  that can be  produced when  89.3  grams  of calcium chloride  reacted  with  excess  sodium phosphate is
   83.08  grams  of calcium phosphate
  calculation
Step 1: write a balanced chemical equation
3CaCl₂  + 2Na₃Po₄ →6NaCl + Ca₃(PO₄)₂
step 2: find the moles of CaClâ‚‚
moles  = mass÷  molar mass
from periodic table the molar mass  of CaCl₂ = 40 +( 2 x 35.5) =111 g/mol
moles = 89.3 g÷111 g/mol = 0.805  moles
Step 3: use the  mole ratio to determine the moles of Ca₃(PO₄)₂
CaCl₂: Ca₃(PO₄)₂  is  3:1 therefore the  moles of Ca₃(PO₄)₂ =0.805  x 1/3 =0.268  moles
Step 4: find mass  of Ca₃(PO₄)₂
mass = moles x  molar mass
from periodic table the molar mass of Ca₃(PO₄)₂ = (40 x3) +[ 31 +(16 x4)]2 =310 g/mol
mass is therefore =0.268 Â moles x 310 g/mol =83.08 grams
