How many grams of copper (II) chloride dihydrate would be requires to react completely with 1.20 g of aluminum

3CuCl2 • 2H2O + Al - - - > 3Cu + 2AlCl3 + 6H2O

moles of Al = 1.2 / 27 = 0.044

1 mole of Al requires 3 moles of CuCl2 • 2H2O

so, 0.044 mole of Al requires = 0.044 x 3 moles of CuCl2 • 2H2O

= 0.133 moles

so mass = 0.133 x 170 = 22.61 g