limiting-reagents-lab-answers

limiting-reagents-lab-answers - CaCl 2 2H 2 O x # g CaCO 3...

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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) 1 mol CaCl 2 •2H 2 O 147.02 g CaCl 2 •2H 2 O x # mol CaCl 2 •2H 2 O = 2.00 g CaCl 2 •2H 2 O 0.0136 mol CaCl 2 •2H 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 2 •2H 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 1 mol Limiting
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4)CaCl 2 •2H 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 1.10 27 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
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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
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Unformatted text preview: CaCl 2 2H 2 O x # g CaCO 3 = 0.0136 mol CaCl 2 2H 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. 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 2 2H 2 O x # g CaCO 3 = 0.0136 mol CaCl 2 2H 2 O 1.36 g CaCO 3 = 100.09 g CaCO 3 1 mol CaCO 3 x or 1 mol CaCl 2 2H 2 O 147.02 g CaCl 2 2H 2 O x 2.00 g CaCl 2 2H 2 O For more lessons, visit www.chalkbored.com...
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This note was uploaded on 11/09/2011 for the course CHEM 110 taught by Professor Sullivan during the Fall '10 term at BYU.

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limiting-reagents-lab-answers - CaCl 2 2H 2 O x # g CaCO 3...

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