The amount of Al that would be needed will be 0.79 grams
Stoichiometric calculations
From the equation of the reaction below:
[tex]2 Al + 3 CuSO_4 --- > Al_2(SO_4)_3 + 3 Cu[/tex]
The mole ratio of Al to [tex]CuSO_4[/tex] is 2:3.
Mole of 71.8 mL. 0.610 M [tex]CuSO_4[/tex] = 0.610 x 71.8/1000 = 0.0438 moles
Equivalent mole of Al = 2/3 x 0.0438 = 0.029 moles
Mass of o.o29 moles Al = 0.029 x 26.98 = 0.79 grams
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