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 CaCl₂
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 moles x 310 g/mol =83.08 grams