Given the following reaction:
Zn + CuCl2 —> ZnCl2 + Cu

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

The answer to your question is 0.35 moles of ZnCl₂

Explanation:

Data

moles of ZnCl₂ = ?

mass of Zn = 23 g

mass of CuCl₂ = excess

Balanced chemical reaction

             Zn + CuCl₂   ⇒   ZnCl₂  +  Cu

Process

1.- Convert the mass of Zn to moles

-Look for the atomic mass of Zn

Atomic mass = 65.4 g

-Convert the grams to moles

            65.4 g 1 mol

             23 g    x

               x = (23 x 1)/65.4

               x = 23 / 65.4

               x = 0.35 moles of Zn

2.- Calculate the moles of ZnCl₂ using proportions and cross multiplication, and the coefficients of the balanced chemical reaction.

              1 mol of Zn 1 mol of ZnCl₂

             0.35 moles of Zn  x

              x = (0.35 x 1) / 1

              x = 0.35 moles of ZnCl₂

a:

by the combination of an unlimited number of elements