Question

If I have 35 grams of aluminum, Al and 29 grams of favonia oxygen gas which will be the limiting reactant?

Answer 1

I have absolutely no idea what "favonia" means, let alone favonia oxygen, so I'll just solve the problem with "regular" oxygen :)

In this case, oxygen will be the limiting reagent. Let's start with the balanced chemical equation

`4Al + 3O_2 -> 2Al_2O_3`

Notice that we have a `4:3` mole-to-mole ratio between `Al` and `O_2`; that is, for every 4 moles of `Al`, at least 3 moles of `O_2` are needed for the formation of 2 moles of `Al_2O_3`.

Knowing `Al` and `O_2`'s molar masses (`27.0 g/(mol)` and `32.0g/(mol)`, respectively), we can determine the number of moles given

`n_(Al) = m_(Al)/(molar.mass) = (35g)/(27.0g/(mol)) = 1.3` moles
`n_(O_2) = m_(O_2)/(molar.mass) = (29g)/(32g/(mol)) = 0.91` moles

Now, let's use the mole-to-mole ratio to determine if `O_2` is the limiting reagent

`1.3 mol esAl * (3mol esO_2)/(4 mol es Al) = 0.98` moles `O_2` required

Since the calculated number of moles of `O_2` is smaller than the required one, oxygen is the limiting reagent.