Pre Lab 8

Topics: Chemistry, Chemical reaction, Concentration Pages: 3 (561 words) Published: April 22, 2015
McKenzie Sluder
Chem 125
Pre Lab Experiment 8
Detminination of % composition of Pennies Using Redox and Double Displacement (Precipitation) Reactions Objections: To determine the percent composition of pennies using oxidation reduction double displacement reactions and titration techniques 1. What Is the weight of a post 1982 penny?

2.5 grams.
2. What is the percent copper and zinc in a post 1982 penny?
The metal composition of the cent 97.5 percent copper and 2.5 percent zinc. 3. How many grams of copper and zinc are in a post 1982 penny?
2.5 grams.
4. How many moles of copper and zinc are in post 1982 pennies?
(2.43 g Zn x 1 mol Zn) / 63.38 g Zn = 0.037 mol Zn
(0.062 g Cu x 1 mol Cu) / 63.54 g Cu = 0.00097 mol Cu
5.Write a balances reaction of zinc with HCL
Zn + 2HCl  ZnCl2 + H2
6. How many moles of HCL are needed to react completely with all of the zinc in a post 1982 penny?
2.43moleZn * (2molesHCl/1moleZn ratio from reaction)=4.6mole HC 7. In a procedure developed to determine the percent zinc in post 1982 pennies, 50 mL of an HCL solution was used to react (dissolve) all of the zinc in the penny. To ensure complete reaction, the solution contains twice as man moles of HCL that is actually needed. What concentration of HCL should be used?

(0.074 mol HCL / 0.05 L ) =1.48 M
8. In the scenario described in problem 7, what is the amount (in moles) of excess (unreacted) HCL in solution?
(1.312 g HCL x 2 mol HCL) / 35.453 g HCL = 0.074 mol HCL
9. How many moles of NaOH would be needed to completely react with all of the excess HCL determined in problem 8?
HCL (aq) + NaOH (aq) H20 (I) + NaCl (aq)
(0.074 mol HCL x 1 mol NaOH) / 1mol HCL = 0.074 mol NaOH
10. As described in problem 7, a procedure was developed to determine the percent in post 1982 pennies. In that procedure 50 mL of an HCL was used to react (dissolve) all of the zinc in the penny. To ensure complete reaction, the solution contains twice as many moles of HCL that...
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