The balanced chemical equation for the precipitation reaction between sodium carbonate Na2CO3 and calcium chloride CaCl2 is:Na2CO3 aq + CaCl2 aq 2 NaCl aq + CaCO3 s To determine the mass of calcium carbonate CaCO3 that can be formed from 25 grams of sodium carbonate Na2CO3 reacting with excess calcium chloride, we will use stoichiometry.First, we need to find the molar mass of Na2CO3 and CaCO3:Na2CO3: 2 22.99 g/mol + 12.01 g/mol + 3 16.00 g/mol = 105.99 g/molCaCO3: 40.08 g/mol + 12.01 g/mol + 3 16.00 g/mol = 100.09 g/molNow, we will convert the mass of Na2CO3 to moles:25 g Na2CO3 1 mol Na2CO3 / 105.99 g Na2CO3 = 0.236 mol Na2CO3According to the balanced equation, 1 mole of Na2CO3 reacts with 1 mole of CaCO3. Therefore, the moles of CaCO3 produced will be the same as the moles of Na2CO3 reacted:0.236 mol Na2CO3 1 mol CaCO3 / 1 mol Na2CO3 = 0.236 mol CaCO3Finally, we will convert the moles of CaCO3 to grams:0.236 mol CaCO3 100.09 g CaCO3 / 1 mol CaCO3 = 23.62 g CaCO3So, 23.62 grams of calcium carbonate can be formed from 25 grams of sodium carbonate reacting with excess calcium chloride.