Answer:
0.321 mole of ZnCl₂ have been produced from 21 g of Zn
Explanation:
The reaction is this one:
Zn (s) + CuCl₂ (aq) → ZnCl₂ (aq) + Cu(s)
A redox one.
If the CuCl₂ is available in excess, we consider the limiting reactant, Zn.
Molar mass Zn = 65.38 g/m
Mass / Molar mass = Mol
21 g/ 65.38 g/m = 0.321 mole
As ratio is 1:1, I will produce the same amount of mole, so I'll make 0.321 mole of ZnCl₂
