chemistry question

Question

According to the following reaction, how many moles of aluminum oxide will be formed upon the complete reaction of 29.4 grams of iron(III) oxide with excess aluminum?

iron(III) oxide (s) + aluminum (s) aluminum oxide (s) + iron (s)

Answer 1

Hi Ashley,

Iron (III) Oxide has a chemical formula of Fe₂O_3. A good place to start is to convert the given mass to moles:

29.4g Fe₂O₃ * (1mol/159.687g) = 0.184 mol Fe₂O₃.

Next let's figure out the chemical formula of Aluminum Oxide (which is Al₂O₃). Notice the bonding pattern is the same here, so this should be a fairly straightforward conversion! Also, the molar mass of Al₂O₃ is 101.960 g/mol... we'll need that in a moment.

So here's our equation (not including the remaining excess aluminum):

__ Fe₂O_{3 (s)} + exc Al _(s) -> __ Al₂O₃ + ___ Fe _(s)

Balanced:

1 Fe₂O_{3 (s)} + exc Al _(s) -> 1 Al₂O₃ + 2 Fe _(s)

Now with our amount (in moles):

0.184 mol Fe₂O_{3 (s)} + exc Al _(s) -> 0.184 mol Al₂O₃ + 0.368 mol Fe _(s)

Now simply convert 0.184 mol Al₂O₃ back to grams and you'll have your answer! This is the reverse of the first step above. Hope this helps in the future!