The amount of Al required will be 15.77 grams
<h3>Stoichiometric problem</h3>First, the equation of the reaction:
The mole ratio is 2:3.
Mole of 63.0 g of FeO = 63/71.84 = 0.8769 moles
Equivalent moles of Al = 0.8769 x 2/3 = 0.5846 moles
Mass of 0.5846 moles Al = 0.5846 x 26.98 = 15.77 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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