Limiting Reactant Lab
Introduction: During this lab, we found the excess reactant and limiting reactant between Aluminum and Copper (ll) Chloride. Using stoichiometry, we were able to make predictions. Through the reaction 2Al + 2CuCl₂ → 3Cu + 2AlCl₃ we carried out this experiment and determined that the limiting reactant was CuCl₂ and that the aluminum was in excess. We also determined the percentage yield of copper.
Purpose: The purpose of this experiment was to determine the limiting reactant and percentage yield of the reaction that took place between Copper (ll) Chloride and Aluminum.
Hypothesis:
2Al + 3CuCl₂ → 3Cu + 2AlCl₃ m= 0.25g m= 0.51g m=?
GMM= 26.9854 GMM= 134.45 GMM= 63.546 n= 0.00926559 n= 0.00379323 nAl= 2Al x0.003379323 3CuCl =0.0025
Therefore CuCl₂ is limiting.
1:1 ratio m= n x GMM =0.0038 x 63.546 =0.24g (theoretical)
Materials:
0.51g of CuCl₂
0.25g of Aluminum
2 100 mL beakers
Stirring rod
Filter paper
Tweezers
Procedure:
1. To begin the reaction, add about 50mL of water to the beaker that contains the aluminum foil and copper (ll) chloride.
2. Record the colour of the solution and any metal that is present at the beginning of the reaction.
3. Record any colour changes as the reaction proceeds. Stir occasionally with the stirring rod.
4. When the reaction is complete, return he beaker, with its contents, to your teacher for proper disposal. Do not pour anything down the drain.
5. Set up the filtration apparatus to collect your Cu from the AlCl₂. solution. Find the mass of your dry paper before you filter.
6. Remove Al from the filter paper, carefully rinsing them with water to prevent loss of Cu.
7. Dry Cu overnight and then find the mass when dry.
Observations:
Material
Before
After
CuCl₂
-blue powder
-crystalline solid
Purpose: The purpose of this experiment was to determine the limiting reactant and percentage yield of the reaction that took place between Copper (ll) Chloride and Aluminum.
Hypothesis:
2Al + 3CuCl₂ → 3Cu + 2AlCl₃ m= 0.25g m= 0.51g m=?
GMM= 26.9854 GMM= 134.45 GMM= 63.546 n= 0.00926559 n= 0.00379323 nAl= 2Al x0.003379323 3CuCl =0.0025
Therefore CuCl₂ is limiting.
1:1 ratio m= n x GMM =0.0038 x 63.546 =0.24g (theoretical)
Materials:
0.51g of CuCl₂
0.25g of Aluminum
2 100 mL beakers
Stirring rod
Filter paper
Tweezers
Procedure:
1. To begin the reaction, add about 50mL of water to the beaker that contains the aluminum foil and copper (ll) chloride.
2. Record the colour of the solution and any metal that is present at the beginning of the reaction.
3. Record any colour changes as the reaction proceeds. Stir occasionally with the stirring rod.
4. When the reaction is complete, return he beaker, with its contents, to your teacher for proper disposal. Do not pour anything down the drain.
5. Set up the filtration apparatus to collect your Cu from the AlCl₂. solution. Find the mass of your dry paper before you filter.
6. Remove Al from the filter paper, carefully rinsing them with water to prevent loss of Cu.
7. Dry Cu overnight and then find the mass when dry.
Observations:
Material
Before
After
CuCl₂
-blue powder
-crystalline solid