# Chemistry: Displacement Reactions

Objectives:
1. Learn how to determinate the percent composition

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2. Learn how to get the percent using oxidation reduction and double reactions

3. To become more familiar with the use titration techniques

4. To learn how to get the salt out of an quimical

Background:
To develop and utilize procedures to determine the percent composition, of ZnCI2. As well titrating with NaOH solution. After all the experiment we got hta salt witch we weighted and we got our results.

Experiment reactions:

.33w of Zn(OH)2 x moles Zn(OH)2/m. moles Zn(OH2) x 1molZnCI2/1 mol Zn(OH)2 x g molZnCI2/1 mol of ZnCL2 = .4531 grams of ZnCI2 .4531g of ZnCl2 x 1 moles ZnCI2/m. moles ZnCI2 x 1molZn/1 mol of ZnCI2 x 65.39 of Zn/1 mol Zn = .2174g Zn

Materials needed:

Scale
Burette
Beaker
Spatula
Prenolpthen (color indicator)
ZnCL2
NaOH
Tube
Vacuum
Paper filters
Rubber base

Questions:
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?

97.5% of zinc and 2.5% of copper
3. How many grams of copper and zinc are in a post 1982 penny?

2.44grams of copper and 0.0625gramss of zinc
4. How many moles of copper and zinc are in post 1982 pennies?

0.0383moles of cooper and 0.000955moles of zinc
5. Write a balanced reaction of zinc with HCl.

Zn (s) + 2 HCl(aq) ====> Zn+2(aq) + 2 Cl-(aq) + H2(g)

Reaction:
Zn(s) + 2 H+(aq) ====> Zn+2(aq) + H2(g)
6. How many moles of HCl are needed to react completely with all of the zinc in a post 1982 penny?
7.46 x 10-2 mol HCl

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 many moles of HCl that is actually needed. What concentration of HCl should be used?

M=2.984
7.46×10-2 mol x 2 = .1492mol H

8. In the scenario described in problem 7, what is the amount (in moles) of excess (unreacted) HCl in solution?
.1492 moles used
.0746 needed
.0746 moles HCI unreacted

9. How many moles of NaOH would be needed to completely react with all of the excess HCl determined in problem 8?
.0746 moles HCI reacts with
.0746 moles of NaOH

10. As described in problem 7, a procedure was developed to determine the percent zinc 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 is actually needed. To determine the percent zinc in the penny, the excess (unreacted) HCl was titrated with NaOH. Determine the concentration of NaOH needed if you want to use approximately 25 mL of NaOH to titrate the excess HCl. M = .0746 moles of NaOH/.025L NaOH

M = 2.984

11. Write the balanced chemical reaction of zinc with HCl (same as problem 5). Is the product of this reaction soluble in aqueous solution? Yes soluble Zn(s) + 2HCI (aq) = ZnCI2(aq) + H2(g)

12. Write the balanced chemical reaction of the product of the reaction described above (problem 11) with NaOH. Is the product of this reaction soluble in aqueous solution? ZnCI2 + 2NaOH = Zn(OH)2+ 2NaCI

Zn(OH)2 = low solubility

Expected results:
Our expected results are to obtain the pink cluster in the solution while titrating. This way after taken more steps we while maybe be able to find some salt.

Steps to do:
*In the first part of this experiment we are going to dissolve the zinc core of a penny *After that we leave the copper covering intact

*We also by putting four notches in the coin using a triangular file and placing the penny in 50ml of a predetermined concentration of HCI overnight

*After that we are going to determinate the concentration needed to add 2 times the number of moles of HCI needed in the 50ml of solution *

We are going to determinate the percent of copper from the mass of copper and the percent zinc by two methods:

1. The titration experiment

2. The precipitation experiment
*Now we first got 10 ml of ZnCI2

*To that we added 6 drops of our color indicator (Prenolpthen) *After that we got 50 ml NaOH solution

*Than we titrate the ZnCl2 with NaOH

*Until we got pink cluster

*Now we put the clusters in a paper filter that was dripping the solution living only the clusters *With a tube we connected a vacuum so this way we were left with only the salt

*After this we put the salt in like an oven so that this could dry the salt and this won’t be wet

*When we took it out all we had was dry salt and we weight it and started our formulas

Results:

I calculated the percent % of salt of Zn(OH)2. First I determinate the formulas for both, follow by calculating the atomic mass of Zn. First I had to have the weight of the salt, so I did all the procedures on top. My result for the weight of the salt was .33g and the result of the Zn was .2174. To be able to get percent I calculated by dividing 2.174 witch is the grams of Zn and 2.5 grams multiplying the result by 100 = 86.994%

Conclusion and discussion:

During Experiment 8 conducted April 16 2014. I noticed the difference in the % for the reaction, when titrating ZnCI2 with NaOH solution. For this experiment we did 2 trials since in the first one we did not get any clusters, but on the second trial we did get some and we got more salt than anyone else.