Name:  _________________________
	Laboratory Activity 9 quarter 3 packet # 6
		Percentage Yield Problem: What is the percentage yield that occurs when Copper (II) Sulfide reacts with Iron? Procedure:    1.) mass out 8 grams of Copper (II) Sulfate in a 250 mL beaker                        2.) mass out 50 mL of distilled water and add to the beaker with CuSO4                       3.) heat the solution to 90 degrees celsius. Do not allow solution to boil                       4.) mass out 2.24 grams of Fe                       5.) when instructed add the iron to the CuSO4 solution very slowly.                       6.) allow beaker to cool and decant                       7.) wash Cu metal with water and decant multiple times                       8.) mass a paper towel carefully then pour the Cu metal onto towel                       9.) allow to dry and weigh again Picture: refer to set-up Chemical Reaction:        Copper (II) Sulfate + Iron -> Iron (II) Sulfate +Copper                  complete:        CuSO4 (aq) + Fe (s) -> Fe SO4 (aq) + Cu(s)                          ionic:      Cu ion(aq) + SO4 ion(aq) + Fe (s) -> Fe ion(aq)+ SO4 ion(aq) + Cu(s)                             net:      Cu ion(aq) + Fe(s) -> Fe ion(aq) + Cu(s) Data:         paper towel + Copper = ________g                   paper towel = _______g                  Copper = _______g Formula:             % yield= actual yield / theoretical yield  x 100                            Safety Notes:   1.) wear safety glasses                         2.) do not burn yourself Conclusion:   what was your actual yield?                       what was the limiting reactant?                       what was the excess reactant?                       what was the theoretical yield?                       what was the percentage yield?