Lag Prelaboratory Assignment Answer the Prelaboratory Questions (these are to be
ID: 712754 • Letter: L
Question
Lag Prelaboratory Assignment Answer the Prelaboratory Questions (these are to be turned in before beginning the lab). It is rec- ommended that you have an additional copy of our Prelaboratory answers to assist you in your lab write-up Write out and balance the possible reactions you will carry out in this lab. 2. 1. If iron(III) sulfate were formed, what theoretical mass of copper is expected? 3. If iron(II) sulfate were formed, what theoretical mass of copper is expected? 1. Place 7.00 g of copper(II) sulfate in a 250-mL beaker. 2. Add about 50 mL of water to the beaker. 3. Place the beaker on a hot plate. 4. Carefully heat and stir the mixture in the beaker. The solution should be hot, but not boiling. After all of the crystals have dissolved, remove the beaker from the heat. Add 2.00 g of iron filings slowly to the hot copper sulfate solution while stirring. Record 5. 6. 7. 8. observations. Allow the beaker to cool for 10-15 minutes. Use suction filtration to collect the copper. Wash the copper with water. Allow the copper to dry. Record the mass of the copper.Explanation / Answer
1) (i) 2 Fe (s) + 3 CuSO4 (aq) -------> Fe2(SO4)3 (aq) + 3 Cu (s)
2 Fe (s) + 3 Cu+2 (aq) -------> 2 Fe3+ (aq) + 3 Cu (s)
(ii) Fe (s) + CuSO4 (aq) ----> FeSO4 (aq) + Cu (s)
Fe (s) + Cu+2 (aq) ----> Fe2+ (aq) + Cu (s)
2) 2 Fe (s) + 3 CuSO4 (aq) -------> Fe2(SO4)3 (aq) + 3 Cu (s)
mol of Fe = 2 gm / 55.8452 gm/mol = 0.03581 mol
Here Fe is the limiting reactant hence,
2 mol of Fe gives 3 mol of Cu when Fe(III) sulfate is formed
0.03581 mol of Fe gives ? mol of Cu when Fe (III) sulfate is formed
(0.03581 * 3) / 2 = 0.053719 mol of Cu is formed
Theoretical yield of Cu = 0.053719 mol * 159.609 g/mol = 8.57403 gms of Cu are formed
3) Fe (s) + CuSO4 (aq) ----> FeSO4 (aq) + Cu (s)
1 mol of Fe gives 1 mol of Cu when Fe(II) sulfate is formed
0.03581 mol of Fe gives ? mol of Cu when Fe(II) sulfate is formed
0.03581 mol of Cu is formed
theoretical yield of Cu = 0.03581 mol * 159.609 gm/mol = 5.7155 gms of Cu are formed
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