First, let's balance the chemical equation for the reaction between calcium chloride CaCl2 and sodium carbonate Na2CO3 :CaCl2 + Na2CO3 CaCO3 + 2NaClNow that the equation is balanced, we can use stoichiometry to determine the mass of the product formed calcium carbonate, CaCO3 when 5.0 grams of calcium chloride react completely with excess sodium carbonate.First, we need to find the molar mass of calcium chloride CaCl2 :Ca = 40.08 g/molCl = 35.45 g/molMolar mass of CaCl2 = 40.08 + 2 35.45 = 111.0 g/molNow, we can convert the mass of calcium chloride to moles:5.0 g CaCl2 1 mol CaCl2 / 111.0 g CaCl2 = 0.0450 mol CaCl2From the balanced chemical equation, we can see that 1 mole of CaCl2 reacts to produce 1 mole of CaCO3. Therefore, the moles of CaCO3 produced will be the same as the moles of CaCl2 reacted:0.0450 mol CaCl2 1 mol CaCO3 / 1 mol CaCl2 = 0.0450 mol CaCO3Now, we need to find the molar mass of calcium carbonate CaCO3 :Ca = 40.08 g/molC = 12.01 g/molO = 16.00 g/molMolar mass of CaCO3 = 40.08 + 12.01 + 3 16.00 = 100.09 g/molFinally, we can convert the moles of CaCO3 produced to grams:0.0450 mol CaCO3 100.09 g CaCO3 / 1 mol CaCO3 = 4.50 g CaCO3So, the mass of the product calcium carbonate formed when 5.0 grams of calcium chloride react completely with excess sodium carbonate is 4.50 grams.