1. Mass of CuCl2 2H2O Beaker 1) .50g Beaker 2) .70g 2. Mass of Al (s) Beaker 1)
ID: 790343 • Letter: 1
Question
1. Mass of CuCl2 2H2O
Beaker 1) .50g
Beaker 2) .70g
2. Mass of Al (s)
Beaker 1) .25g
Beaker 2) .05 g
3. Mass of filter paper = .77g
4. mass of filter paper + contents from beaker 2 = 3.49
5. Mass of CuCl2 2H2O
Beaker 3) .70
6. Mass of Al (s)
Beaker 3) .07
Moles of AlCl3 produced (theoretical) = ?
8. Beaker 2 calculations
Moles of CuCl2 2H2O =?
Moles of Al (s) =?
Moles of Cu (s) produced (theoretical) =?
Moles of AlCl3 produced (theoretical) = ?
Mass of Cu (s) produced (acutal) = 2.72
Moles of Cu(s) produced (acutual) = ?
Moles of AlCl3 produced (theoretical) = ?
Explanation / Answer
Beaker 1:
Mass of CuCl2 2H2O = 0.5 gms
Molecular mass of CuCl2 2H2O = 170.48 gms
No. of moles of CuCl2 2H2O = 0.5/170.48 = 2.933 x 10^-3 moles
Mass of Al = 0.7 gms
Molecular mass of Al = 27 gms
No. of moles of Al = 25.925 x 10^-3
3CuCl2 2H2O + 2Al --> 2AlCl3 + 3Cu + 6H2O
Here Al is in excess. So, no. of moles of Cu(s) produced is equal to 2.933 x 10^-3 moles as Al is in excess and all the CuCl2 2H2O reacts.
From the balanced reaction, for every 3 moles of CuCl2 2H2O, 2 moles of AlCl3 is produced. Hence, No. of moles of AlCl3 produced = 2/3 x 2.933 x 10^-3 = 1.955 x 10^-3 moles.
Beaker 2:
Mass of CuCl2 2H2O = 0.7 gms
Molecular mass of CuCl2 2H2O = 170.48 gms
No. of moles of CuCl2 2H2O = 0.7/170.48 = 4.106 x 10^-3 moles
Mass of Al = 0.05 gms
Molecular mass of Al = 27 gms
No. of moles of Al = 1.852 x 10^-3
3CuCl2 2H2O + 2Al --> 2AlCl3 + 3Cu + 6H2O
Al required for total reaction = 2/3 x 4.106 x 10^-3 = 2.737 x 10^-3 moles
Al present in the beaker = 1.852 x 10^-3 moles
So CuCl2 2H2O is in excess. All the aluminium reacts.
Moles of AlCl3 produced theoretically = 1.852 x 10^-3 moles
Moles of Cu(s) produced theoretically = 3/2 x 1.852 x 10^-3 = 2.778 x10^-3 moles
Mass of Cu(s) actually produced = 2.72 gms
Moles of Cu(s) actually produced = 42.834 x 10^-3 moles
Beaker 3:
Mass of CuCl2 2H2O = 0.7 gms
Molecular mass of CuCl2 2H2O = 170.48 gms
No. of moles of CuCl2 2H2O = 0.7/170.48 = 4.106 x 10^-3 moles
Mass of Al = 0.07 gms
Molecular mass of Al = 27 gms
No. of moles of Al = 2.596 x 10^-3 moles
(3/2 x No. of moles of Al) moles of CuCl2 2H2O is required for compete reaction theoretically which is 3.894 x 10^-3 moles. No of moles present is 4.106 x 10^-3. So CuCl2 2H2O is in excess and all the Al reacts.
No. of moles of AlCl3 formed = 2.596 x 10^-3 moles
No. of moles of Cu(s) formed = 3.894 x 10^-3 moles
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