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The amount of Al that would be needed will be 0.79 grams
<h3>Stoichiometric calculations</h3>From the equation of the reaction below:
The mole ratio of Al to is 2:3.
Mole of 71.8 mL. 0.610 M = 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: brainly.com/question/27287858
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