Iron(III) oxide, Fe2O3, is a result of the reaction of iron with the oxygen in air. a. What is the balanced equation for this reaction? (Use the lowest possible coefficients. Omit states of matter.) b. What number of moles of iron reacts with 17.15 mol of oxygen from the air? mol c. What mass of iron is required to react with 17.15 mol of oxygen? Mass =

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Ondinne Ondinne · Answer 1 · 2019-10-03T17:30:20+02:00

Answer:

(a) [tex]_{4}Fe+_{3}O_{2}=_{2}Fe_{2}O_{3}[/tex]

(b) 22.9 moles of Fe

(c) 1278.8 g of Fe

Explanation:

(a) To write the equation you have to take in account that as the problem says that is oxygen in air you should write it as [tex]O_{2}[/tex]

So, first the problem is telling you that iron reacts with oxygen in air, so you should write the reactants:

[tex]Fe+O_{2}[/tex]

Then the problem says that it produces iron(III) oxide [tex]Fe_{2}O_{3}[/tex], so you should write the product:

[tex]Fe+O_{2}=Fe_{2}O_{3}[/tex]

Finally you should balance the equation, so you need to have the same quantity of each atom in both sides of the equation:

[tex]_{4}Fe+_{3}O_{2}=_{2}Fe_{2}O_{3}[/tex]

(b) You should use the stoichiometry of the balanced equation to solve this part, so:

[tex]17.15molesO_{2}*\frac{4molesFe}{3molesO_{2}}=22.9molesFe[/tex]

(c) To solve this part, you need to know that the molar mass for the Fe is [tex]55.845\frac{g}{mol}[/tex], so:

[tex]22.9molesFe*\frac{55.845gFe}{1molFe}=1278.8gFe[/tex]