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Mass Relationships in a Chemical Reaction – Lab

Christian Lecce
Mr. Ribarich
Wednesday, February 20st, 2013

Purpose

To determine the mass of copper formed when excess aluminum is reacted with a given mass of a copper salt (Copper Chloride dihydrate), and the mole-to-mole ratio between the reactant and the product of a chemical reaction.

Apparatus

* 150ml beaker
* Stirring rod
* Ruler
* Hotplate
* Tweezers
* 50ml graduated cylinder

Materials

* Copper (II) chloride dehydrate
* Aluminum foil

Procedure

1) Determine the mass of a clean 150ml beaker.

2) Add 2g of copper (II) chloride dehydrate

3) Add 50ml of water and stir to dissolve the compound

4) Fold a 6cm x 10cm strip of aluminum foil lengthwise twice, then curl the folded strip into a circle so it fits on its edge in the beaker

5) Heat the solution with the aluminum until the solution loses its blue colour and turns brown with copper.

6) Once no more reaction can occur with the aluminum, remove the excess with out removing any copper

7) Heat the content until all of the liquid evaporates and there is no more smoke emitting from the beaker.

8) Once the beaker has nothing but copper in it, weigh it.

9) Subtract the total weight by the original weight of the beaker to get the weight of just the copper

Observations

1) The theoretical yield comes to 0.7435g of copper

3CuCl2-2H2O + 2Al -> 3Cu + 2AlCl3 + 6H2O
Mass of CuCl2-2H2O| Mass of Clean Beaker| Mass of Copper created| Mole-to-Mole ratio| 2g| 113.5g| 0.8g| 1 : 1 (hydrate to copper)|

2g of CuCl2-2H2O CuCl2-2H2O = 170.55 g/mol
2 / 170.55 = 0.0117mol

1 : 1
0.0117 x

0.0117 moles of Cu
Cu = 63.55g/mol
63.55 x 0.0117 = 0.7435g

2) The actual yield of the reaction is 0.8
0.8 - 0.7425 = 0.0575g is the difference

3) The percentage yield is: 107.5%

Actual / Theoretical
0.8 / 0.7435= 107.5

4) 0.2106g of aluminum will be needed to react with 2g of copper chloride dihydrate

3CuCl2-2H2O + 2Al -> 3Cu + 2AlCl3 + 6H2O

CuCl2-2H2O = 170.55g/mol
2 / 170.55 = 0.0117mol

3 : 2 3x = 0.0234 0.0117 x x = 0.0078

0.0078mol of Al
Al = 27g/mol
0.0078 x 27 = 0.2106g

5) 0.0117 moles of copper chloride dihydrate were supposed to be used, as well as 0.0117 moles of copper created.

6) The mole to mole ratio is 1 : 1. When 1 mole of copper chloride dihydrate reacts, 1 mole of copper should be produced.

7) The ratio obtained agrees with the balanced chemical equation because 3 : 3 in the equation is equal to 1 : 1.

8) If less than the correct amount were used then there would not be as high a yield for copper in the product.

9) If the copper that was produced were heated further until it turned black, then the weight of the copper would change. Some would be eliminated.

10) No, aluminum is more reactive than copper, therefore copper could not displace aluminum when it is in a compound to create a reaction. 11) If 3g of copper chloride dihydrate reacted with 3g (excess) aluminum, then 1.545g of aluminum chloride could be formed

3g of CuCl2-2H2O 3g of Al
CuCl2-2H2O = 170.55g/mol Al = 27g/mol
3 / 170.55 = 0.0175 mol 3 / 27 = 0.1111
* LIMITING * * EXCESS *

3 : 2 3x = 0.035
0.0175 x x = 0.0116

0.0116 mol of AlCl3
AlCl3 = 133.2g/mol
0.0116 x 133.2 = 1.545g

Conclusion

As stated in the observations, the mass of copper formed was 0.8 grams. The mole-to-mole ratio between the reactant (CuCl2 – 2H2O) and the product (Cu) was 1:1.

Sources of Error

* Pieces...
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