Pre Lab 8
November 3, 2014
Chemistry 1411- 106
Pre Lab
Experiment #8
Objective: This Lab will help us to understand oxidation reduction and double displacement through finding the percent composition of pennies. We will also once again be working with titration in this lab. We will titrate the solution until we are only left with a solid form which will tell us about the composition of pennies.
Introduction: The weight of a post 1982 penny is 2.5 grams, and the percent of zinc is 97.5% leaving only 2.5% copper. The amount of Zinc in a post 1982 penny is 2.437 grams, and the amount of copper is 0.062 grams. Along with all of this we must know the amount of moles of each. To find the amount of moles we must solve for them:
Zinc = (2.4375g / 1) X (1 / 65.38g) = 3.73 X 10-2 moles
Copper = (0.0625g / 1) X (1 / 63.546g) = 9.835 X 10-4 moles
The balanced equation for zinc and HCl is Zn + 2HCl = 1H2 + 1 ZnCl2
H=1, Zn = 65.38, Cl=35.45
Molar Mass of HCl = 101.83
(3.73 X 10-2 moles Zn) (2HCl/1Zn) = 7.46 X 10-2 moles HCl
The amount of HCl that should be used is:
50mL of HCl = .05 L of HCl
7.46 X 10-2 moles of HCl X 2 = .1492 moles of HCl used and the molarity would be .1492 moles HCl/ .05 liters HCl = 2.984 molarity. The amount of unreacted HCl in the solution is shown here: 0.1492 moles used - .0746 moles needed = 0.0746 moles unreacted. The amount of NaOH moles needed to completely react with this excess HCl is .0746 moles of HCl and .0746 moles of NaOH. To titrate the excess HCl you would need a molarity of 2.984 NaOH. In the equation Zn(s) + 2HCl (aq) = H2 (g) + ZnCl2 (aq) ZnCl2 is soluble. When using NaOH the equation would be ZnCl2 + 2NaOH = Zn (OH)2 + ZnCl and Zn(OH)2 has a low solubility.
Procedure:
In this part of the lab we will be titrating ZnCl2 and NaOH with HCl to figure out how much zinc is in a penny. We will titrate these until we get a pink indicator color. Once the pink indicator color is shown, there will also be a solid form of Zn(OH)2. Once you get the
Chemistry 1411- 106
Pre Lab
Experiment #8
Objective: This Lab will help us to understand oxidation reduction and double displacement through finding the percent composition of pennies. We will also once again be working with titration in this lab. We will titrate the solution until we are only left with a solid form which will tell us about the composition of pennies.
Introduction: The weight of a post 1982 penny is 2.5 grams, and the percent of zinc is 97.5% leaving only 2.5% copper. The amount of Zinc in a post 1982 penny is 2.437 grams, and the amount of copper is 0.062 grams. Along with all of this we must know the amount of moles of each. To find the amount of moles we must solve for them:
Zinc = (2.4375g / 1) X (1 / 65.38g) = 3.73 X 10-2 moles
Copper = (0.0625g / 1) X (1 / 63.546g) = 9.835 X 10-4 moles
The balanced equation for zinc and HCl is Zn + 2HCl = 1H2 + 1 ZnCl2
H=1, Zn = 65.38, Cl=35.45
Molar Mass of HCl = 101.83
(3.73 X 10-2 moles Zn) (2HCl/1Zn) = 7.46 X 10-2 moles HCl
The amount of HCl that should be used is:
50mL of HCl = .05 L of HCl
7.46 X 10-2 moles of HCl X 2 = .1492 moles of HCl used and the molarity would be .1492 moles HCl/ .05 liters HCl = 2.984 molarity. The amount of unreacted HCl in the solution is shown here: 0.1492 moles used - .0746 moles needed = 0.0746 moles unreacted. The amount of NaOH moles needed to completely react with this excess HCl is .0746 moles of HCl and .0746 moles of NaOH. To titrate the excess HCl you would need a molarity of 2.984 NaOH. In the equation Zn(s) + 2HCl (aq) = H2 (g) + ZnCl2 (aq) ZnCl2 is soluble. When using NaOH the equation would be ZnCl2 + 2NaOH = Zn (OH)2 + ZnCl and Zn(OH)2 has a low solubility.
Procedure:
In this part of the lab we will be titrating ZnCl2 and NaOH with HCl to figure out how much zinc is in a penny. We will titrate these until we get a pink indicator color. Once the pink indicator color is shown, there will also be a solid form of Zn(OH)2. Once you get the