2Al + 3CuClâ‚‚ â†’ 3Cu + 2AlClâ‚ƒ 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
Therefore CuClâ‚‚ is limiting.
m= n x GMM
=0.0038 x 63.546
0.51g of CuClâ‚‚
0.25g of Aluminum
2 100 mL beakers
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 AlClâ‚‚. 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.
-black precipitate formed on the tiny pieces of aluminum
Actual yield 0.20g
Theoretical yield 0.24g
Percentage yield = AY
1. According to our observations, the aluminum was in excess, there was bits
of it floating around after the CuClâ‚‚ was gone. That means the CuClâ‚‚ was limiting. 2. My prediction matches up with what i predicted.
4. My prediction was correct.
In this experiment, we found that the limiting reactant was CuClâ‚‚, which was the same as our