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1) 2) CaCl 2 • 2H 2 O(aq) + Na 2 CO 3 (aq)  2H 2 O + CaCO 3 (s) + 2NaCl(aq) 3)

0.00943 mol Na 2 CO 3. = . 1 mol CaCl 2 • 2H 2 O. 0.0136 mol CaCl 2 • 2H 2 O. x . = . 147.02 g CaCl 2 • 2H 2 O. 1 mol Na 2 CO 3. x . 105.99 g Na 2 CO 3. # mol CaCl 2 • 2H 2 O =. 2.00 g CaCl 2 • 2H 2 O. 1) 2) CaCl 2 • 2H 2 O(aq) + Na 2 CO 3 (aq)

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1) 2) CaCl 2 • 2H 2 O(aq) + Na 2 CO 3 (aq)  2H 2 O + CaCO 3 (s) + 2NaCl(aq) 3)

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  1. 0.00943 mol Na2CO3 = 1 mol CaCl2•2H2O 0.0136 mol CaCl2•2H2O x = 147.02 g CaCl2•2H2O 1 mol Na2CO3 x 105.99 g Na2CO3 # mol CaCl2•2H2O= 2.00 g CaCl2•2H2O 1) 2) CaCl2•2H2O(aq) + Na2CO3(aq)  2H2O + CaCO3(s) + 2NaCl(aq) 3) # mol Na2CO3= 1.00 g Na2CO3 0.0136 mol 0.00943 mol .0136/.00943 = 1.44 mol .00943/.00943 = 1 Limiting 1 mol 1 mol

  2. 58.44 g NaCl x 2 mol NaCl 1 mol NaCl x 1 mol Na2CO3 or = 1.1027 g NaCl 1.00 g Na2CO3 1 mol Na2CO3 x 105.99 g Na2CO3 • CaCl2•2H2O is in excess. It is aqueous, so it will stay dissolved in water, ending up in the filtrate. Adding Na2CO3 to the filtrate will make it cloudy, revealing the presence of CaCl2. # g NaCl = 0.00943 mol Na2CO3

  3. 1 mol CaCO3 1 mol CaCO3 100.09 g CaCO3 100.09 g CaCO3 x x x x 1 mol Na2CO3 1 mol CaCl2•2H2O 1 mol CaCO3 1 mol CaCO3 = = 0.944 g CaCO3 1.36 g CaCO3 # g CaCO3 = 0.00943 mol Na2CO3 # g CaCO3 = 0.0136 mol CaCl2•2H2O • Hopefully the prediction was correct (closer to 6). The comparison of 6 vs. 7 gives us a shortcut for determining limiting reagents.

  4. 1 mol CaCO3 1 mol CaCO3 100.09 g CaCO3 100.09 g CaCO3 x x x x 1 mol Na2CO3 1 mol CaCl2•2H2O 1 mol CaCO3 1 mol CaCO3 or 1.00 g Na2CO3 1 mol CaCl2•2H2O x 147.02 g CaCl2•2H2O = = 1.36 g CaCO3 0.944 g CaCO3 or 1 mol Na2CO3 x 2.00 g CaCl2•2H2O 105.99 g Na2CO3 For more lessons, visit www.chalkbored.com # g CaCO3 = 0.00943 mol Na2CO3 # g CaCO3 = 0.0136 mol CaCl2•2H2O

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