1. 1) 2) CaCl 22H 2 O(aq) + Na 2 CO 3 (aq) 2H 2 O + CaCO 3 (s) + 2NaCl(aq) 3) 1 mol CaCl 22H 2 O 147.02 g CaCl 22H 2 O x # mol CaCl 22H 2 O = 2.00 g CaCl 22H 2 O 0.0136 mol CaCl 22H 2 O = 1 mol Na 2 CO 3 105.99 g Na 2 CO 3 x # mol Na 2 CO 3 = 1.00 g Na 2 CO 3 0.00943 mol Na 2 CO 3 = CaCl 22H 2 O Na 2 CO 3 What we have What we need 0.0136 mol 0.00943 mol.0136/.00943 = 1.44 mol.00943/.00943 = 1 1 mol Limiting

2. 4)CaCl 22H 2 O is in excess. It is aqueous, so it will stay dissolved in water, ending up in the filtrate. Adding Na 2 CO 3 to the filtrate will make it cloudy, revealing the presence of CaCl 2. 5) 2 mol NaCl 1 mol Na 2 CO 3 x # g NaCl = 0.00943 mol Na 2 CO 3 1.1027 g NaCl= 58.44 g NaCl 1 mol NaCl x 1 mol Na 2 CO 3 105.99 g Na 2 CO 3 x 1.00 g Na 2 CO 3 or

3. 6) 1 mol CaCO 3 1 mol Na 2 CO 3 x # g CaCO 3 = 0.00943 mol Na 2 CO 3 0.944 g CaCO 3 = 100.09 g CaCO 3 1 mol CaCO 3 x 7) 1 mol CaCO 3 1 mol CaCl 22H 2 O x # g CaCO 3 = 0.0136 mol CaCl 22H 2 O 1.36 g CaCO 3 = 100.09 g CaCO 3 1 mol CaCO 3 x 8)Hopefully the prediction was correct (closer to 6). The comparison of 6 vs. 7 gives us a shortcut for determining limiting reagents.

4. 6) 1 mol CaCO 3 1 mol Na 2 CO 3 x # g CaCO 3 = 0.00943 mol Na 2 CO 3 0.944 g CaCO 3 = 100.09 g CaCO 3 1 mol CaCO 3 x 1 mol Na 2 CO 3 105.99 g Na 2 CO 3 x 1.00 g Na 2 CO 3 or 7) 1 mol CaCO 3 1 mol CaCl 22H 2 O x # g CaCO 3 = 0.0136 mol CaCl 22H 2 O 1.36 g CaCO 3 = 100.09 g CaCO 3 1 mol CaCO 3 x or 1 mol CaCl 22H 2 O 147.02 g CaCl 22H 2 O x 2.00 g CaCl 22H 2 O For more lessons, visit www.chalkbored.com www.chalkbored.com