Question

How many grams of Al would be needed to displace all of the copper ions in 71.8 mL of 0.610 M CuSO4?

Answer 1

The amount of Al that would be needed will be 0.79 grams

Stoichiometric calculations

From the equation of the reaction below:

$2 Al + 3 CuSO_4 --- > Al_2(SO_4)_3 + 3 Cu$

The mole ratio of Al to $CuSO_4$ is 2:3.

Mole of 71.8 mL. 0.610 M  $CuSO_4$ = 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

More on stoichiometric calculations can be found here: https://brainly.com/question/27287858

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