Answer to Question #71627 in General Chemistry for cassie

Question

We performed a lab that involved the single displacement reaction of copper(II)hydrate and aluminum. How do you calculate the theoretical yield and the actual yield of copper?
This is my data;
Element Mass (g)
CuCl2 ‘2 H2O 18.58g
Beaker 69.43g
Stirring Rod 12.03g
Aluminium Foil 0.55g
Filter Paper 1.03g
Distilled Water 50g
Copper Salt 1.85g

Answer 1

You can find out the theoretical yield in this way:
m(CuCl2*2 H2O)=18.58g
m(Al) = 0.55g
3CuCl2 + 2Al = 2AlCl3 + 3Cu
n(Al) =0.55g / 27g/mol = 0.02mol
w(CuCl2)=m(CuCl2)/ m(CuCl2*2 H2O) = 135g/mol / 171g/mol = 0.79
m(CuCl2)= m(CuCl2*2 H2O)* w(CuCl2)=18.58g*0.79 = 14.67g
n(CuCl2)=m (CuCl2)/ M(CuCl2)=14.67g/ 135g/mol= 0.108mol
n(Al)< n(CuCl2) that’s why we chose n(Al)
n(Cu)=3/2 n(Al)=0.03 mol
m(Cu)= n(Cu)*M(Cu)= 0.03mol *64g/mol = 1.92g
But the actual yield you have (Copper Salt 1.85g), but it has incorrect name.
It must be Copper 1.85g