Answer :

About 45.6 gAL are need to completely react with 135g FE2O3

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To determine the mass of Al needed to react with 135 grams of Fe₂O₃, convert the mass of Fe₂O₃ to moles, use the stoichiometric ratio to find moles of Al needed, and then convert those moles to grams of Al to three significant figures.

The question is asking how many grams of aluminum (Al) are needed to completely react with 135 grams of iron(III) oxide (Fe2O3). This is a stoichiometry problem that requires using the balanced chemical equation:

Fe₂O₃(s) + 2Al(s) →2Fe(s) + Al₂O₃(s)

We start by converting the mass of Fe₂O₃ to moles using its molar mass:

1.0g Fe₂O₃ / 159.7 g/mol = moles of Fe₂O₃

Then we use the mole ratio from the balanced equation to convert from moles of Fe₂O₃ to moles of Al. The molar mass of Al is 27.0 g/mol. Finally, we convert moles of Al to grams using its molar mass to find the mass of Al needed for the reaction.

With the provided mass of 135g of Fe₂O₃, and the reaction stoichiometry, we can determine the mass of Al required which will be expressed to three significant figures.