What is the molarity of ZnCl2 that forms when 20.0 g of zinc completely reacts with CuCl2 according to the following reaction? Assume a final volume of 255 mL. Zn (s) + CuCl2 (aq) →ZnCl2 (aq) + Cu (s)

Molarity ZnCl2 = 1.2 M

Explanation:

Step 1: Data given

Mass of zinc = 20.0 grams

Atomic mass of zinc = 65.38 g/mol

Volume = 255 mL = 0.255 L

Step 2: The balanced equation

Zn (s) + CuCl2 (aq) → ZnCl2 (aq) + Cu (s)

Step 3: Calculate moles Zn

Moles Zn = mass Zn / molar mass Zn

Moles Zn = 20.0 grams / 65.38 g/mol

Moles Zn = 0.306 moles

Step 4: Calculate moles ZnCl2

For 1 mol Zn we need 1 mol CuCl2 to produce 1 mol ZnCl2 and 1 mol Cu

For 0.306 moles Zn we have 0.306 moles ZnCl2

Step 5: Calculate molarity ZnCl2

Molarity = 0.306 moles / 0.255 L

Molarity ZnCl2 = 1.2 M