ZnO + 2HCl - - > ZnCl2 + H2O

what is the maximum amount of ZnCl2 which could be formed from 3.52 mol of ZnO and 18.82 mol of HCl

Theoretical yield is the maximum amount of product that could be formed from amounts of reactants.

Now, in this problem we will find the theoretical yield of ZnCl2

3.52 mol ZnO x (1 mol ZnCl2 / 1 mol ZnO) = 3.52 mol ZnCl2

18.82 mol HCl x (1 mol ZnCl2 / 2 mol HCl) = 9.41 mol ZnCl2

ZnO is the limiting reactant.

Therefore, we form 3.52 mol of ZnCl2

molar mass of ZnCl2 is

1 (65.38) + 2 (35.45) = 135.28 g/mol

This has a mass of

m = M x n

m = 135.28 g/mol ZnCl2 x 3.52 mol ZnCl2

m = 476.2 g ZnCl2

Answer: maximum amount of ZnCl2 is 476.2 g