2Al + 3Cl2 = 2AlCl3
nAl=nAlCl3=5 moli Al
m Al pur = 5*27=135g Al
m Al impur = 135*100/60=225g Al
b) 3moli Cl............2moli AlCl3
x=7,5 moli................5 moli AlCl3
1 mol Cl..........................6,022*10^23 molecule
7,5 moli...........................x=45,165*10^23 molecule