stoichiometry chem questin

Question

How many moles of ZnClX2​ will be produced from 23.0 g23, point, 0, start text, space, g, end text of ZnZ, n, assuming CuClX2​ is available in excess?

Answer 1

Your question is a bit random, but piecing things together I think the question is asking...

How many moles of ZnCl₂ will be produced from 23.0 g of Zn assuming that CuCl₂ is available in excess?

Write a correctly balanced equation:

Zn(s) + CuCl₂(aq) ==> ZnCl₂(aq) + Cu(s)

Find moles of Zn present:

23.0 g Zn x 1 mol Zn/65.39 g = 0.3517 = 0.352 moles Zn

Next find moles of ZnCl₂ formed:

0.3517 moles Zn x 1 mol ZnCl₂/mol Zn = 0.3517 moles = 0.352 moles ZnCl₂ formed

Answer 2

Hello Nevaeh!

If I understood correctly, then the reaction you're looking for is:

CuCl₂ + Zn → ZnCl₂ + Cu

Note that 1 mol of Zn produces 1 mol of ZnCl₂ if there is enough CuCl₂

The molecular weight of Zn is 65.38 g/mol. Then, the number of mols in 23.0 g is:

23/65.38 = 0.352 mols of Zn

Therefore, it will be produced 0.352 mols of Zn*(1 mol of ZnCl₂/1 mol of Zn)= 0.352 mols of ZnCl₂