Question

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

Answer 1

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