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

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

$4 A l + 3 {O}_{2} \to 2 A {l}_{2} {O}_{3}$

Notice that we have a $4 : 3$ mole-to-mole ratio between $A l$ and ${O}_{2}$; that is, for every 4 moles of $A l$, at least 3 moles of ${O}_{2}$ are needed for the formation of 2 moles of $A {l}_{2} {O}_{3}$.

Knowing $A l$ and ${O}_{2}$'s molar masses ($27.0 \frac{g}{m o l}$ and $32.0 \frac{g}{m o l}$, respectively), we can determine the number of moles given

${n}_{A l} = {m}_{A l} / \left(m o l a r . m a s s\right) = \frac{35 g}{27.0 \frac{g}{m o l}} = 1.3$ moles
${n}_{{O}_{2}} = {m}_{{O}_{2}} / \left(m o l a r . m a s s\right) = \frac{29 g}{32 \frac{g}{m o l}} = 0.91$ moles

Now, let's use the mole-to-mole ratio to determine if ${O}_{2}$ is the limiting reagent

$1.3 m o l e s A l \cdot \frac{3 m o l e s {O}_{2}}{4 m o l e s A l} = 0.98$ moles ${O}_{2}$ required

Since the calculated number of moles of ${O}_{2}$ is smaller than the required one, oxygen is the limiting reagent.