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:
So, the amount of CuCl₂ needed to react an entire 35-gram sample of aluminum would be 261,62 g CuCl2
Local or national ISO I will figure it out
If the water is used fo energy then the same present of the water would be the same as the energy.
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Hope this helped