Question

Zn + CuCl2 --> ZnCl2 + Cu. How many moles of ZnCl2 will be produced from 21.0 g of Zn, assuming CuCl2 is available in excess?

a) 0.325 mol
b) 0.375 mol
c) 0.650 mol
d) 0.750 mol

Answer 1

The moles of ZnCl₂ produced from 21.0 g of Zn are calculated by first converting the mass of Zn to moles and then using the stoichiometry of the balanced chemical equation. The closest answer is 0.325 mol of ZnCl₂.

Step-by-step explanation:

To calculate the moles of ZnCl₂ produced from 21.0 g of Zn, we first need to know the molar mass of Zn, which is 65.38 g/mol. The balanced chemical equation Zn (s) + CuCl₂ (s) → ZnCl₂ (s) + Cu (s) shows us a stoichiometry of 1:1 between Zn and ZnCl₂.

First, we convert the mass of Zn to moles by dividing by its molar mass:

21.0 g Zn × (1 mol Zn / 65.38 g Zn) = 0.321 mol Zn

Since the molar ratio of Zn to ZnCl₂ in the reaction is 1:1, this is also the number of moles of ZnCl₂ that would be formed:

0.321 mol Zn → 0.321 mol ZnCl₂

Thus, answer option (a) 0.325 mol is the closest answer to our calculated value.