Answer:
Mass of aluminum required is 1.277 grams
Explanation:
the reaction between copper and aluminium is a redox reaction and this can be written as:
[tex]3Cu^{+2}+2Al(s) ---> 2Al^{+3} +3Cu(s)[/tex]
Thus we need two moles of aluminum to reduce or to produce three moles of copper metal.
The moles of copper need to produce
= [tex]\frac{mass}{atomicmass}=\frac{4.5}{63.5}=0.071mol[/tex]
These will produced from = [tex]\frac{2X0.071}{3}=0.0473mol[/tex] of aluminum
The mass of aluminum required = moles X atomic mass = 0.0473X27=1.277g
