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.

Related questions
See all questions in Limiting Reagent
Impact of this question
1779 views around the world
You can reuse this answer
Creative Commons License