Question

aluminum metal reacts with aqueous copper (II) chloride to forma aluminum chloride & copper metal. how many grams of copper are needed to react an entire 35 gram sample of aluminum

Answer 1

Hi! The reaction is the following one:

2Al + 3CuCl₂ → 2AlCl₃ + 3Cu 

So if you want to know the grams of copper (II) chloride needed (you can't react the copper metal with aluminum metal. It should be the copper (II) chloride), you should use the following conversion factor, using the molar masses of the reagents:

$35 g Al* \frac{1 mol Al}{26,98 g Al} * \frac{3 mol CuCl2}{2 mol Al} * \frac{134,45 g CuCl2}{1 mol CuCl2} = 261,62 g CuCl2$

So, the amount of CuCl₂ needed to react an entire 35-gram sample of aluminum would be 261,62 g CuCl2