What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) - 2Fe(l) + Al_2O_3(s)#?

Question

What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) -> 2Fe(l) + Al_2O_3(s)#?

Wyatt Adkins · Accepted Answer

#13.5g#

#n_(Fe_2O_3)=m/(m_r)=40/(2xx55.8+3xx16)=0.25mol#.

The mole ratio in which chemicals combine is represented by the balanced chemical equation.

Hence 1 mole iron(III)oxide requires 2 moles aluminium. Therefore by ratio and proportion, #0.25 mol# iron(III)oxide requires #0.50# moles aluminium. This corresponds to a mass of :

#m_(Al)=nxxM_r=0.50xx27=13.5g#.

Ellie Andrews · Answer

undefined To determine the mass of aluminum required to completely consume 40.0 grams of iron(III) oxide in the given reaction, we first need to find the molar mass of iron(III) oxide (Fe2O3).

The molar mass of Fe2O3 = (2 * molar mass of Fe) + (3 * molar mass of O) = (2 * 55.845 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol.

Now, we calculate the number of moles of Fe2O3 present in 40.0 grams:

Number of moles = Mass / Molar mass = 40.0 g / 159.69 g/mol ≈ 0.2504 mol.

According to the balanced chemical equation, 1 mole of Fe2O3 reacts with 2 moles of aluminum (Al).

So, the number of moles of aluminum required = 0.2504 mol Fe2O3 * (2 mol Al / 1 mol Fe2O3) = 0.5008 mol.

Now, we find the mass of aluminum required:

Mass = Number of moles * Molar mass of Al = 0.5008 mol * 26.982 g/mol (molar mass of Al) ≈ 13.49 grams.

Therefore, approximately 13.49 grams of aluminum is required to completely consume 40.0 grams of iron(III) oxide in the reaction.

What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) -&gt; 2Fe(l) + Al_2O_3(s)#?

What mass of aluminum is required if 40.0 grams of iron (III) oxide is to be completely consumed in the reaction #Fe_2O_3(s) + 2Al(s) -> 2Fe(l) + Al_2O_3(s)#?