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

1 Answer
Stefan V.
Dec 22, 2014

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.

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