To a solution containing 5.5 g of calcium chloride was added 3.18 g of sodium carbonate.

To a solution containing 5.5 g of calcium chloride was added 3.18 g of sodium carbonate. Calculate the mass of the precipitate formed.

Reaction of calcium chloride and sodium carbonate to form sodium chloride and calcium carbonate as a precipitate.
CaCl2 + Na2CO3 = CaCO3 + 2NaCl.
Let’s determine the number of moles of substances.
n (CaCl2) = m / Mr = 5.5 g / 111 g / mol = 0.05 mol.
n (Na2CO3) = 3.18 g / 106 g / mol = 0.03 mol.
Chloride in excess is calculated using sodium carbonate.
n (CaCO3) = 0.03 mol.
m (CaCO3) = n • Mr = 0.03 mol • 100 g / mol = 3 g.