Solve Stoichiometry: 3.4 moles Aluminum Oxide from Al + O2

[Youngstarr1589]

Aluminum oxide is formed from the reaction of metallic aluminum with oxygen gas. How many moles of Aluminum are needed to form 3.4 moles of Aluminum oxide?
Hint: This is a simple ratio problem. Just use the molar ratio from the balanced equation. This was covered in 5.09. Mass is not involved.

I need a lot of help with this problem... where do I start?

 

[PPonte]

First, it would help to write the chemical equation of the mentioned chemical reaction.

Second, balance the chemical equation.

Third, by the stoichiometric proportion, find the number of moles of alluminium that you would need to form 3.4 moles of aluminum oxide.

Hint: $$\frac{n_{Al}}{2} = \frac{n_{Al_2O_3}}{1}$$

 

[Youngstarr1589]

Ok this is what I have so far... I balanced the equation and got 4Al + O2
--> 2Al2O3 so then I did 3.4 molAl2O3 x 4mol Al = 6.8 mol Al ... is this right?

 

[PPonte]

Yes, you are correct.

 

[Borek]

Funny, exactly this reaction forms basis of the stoichiometry calculations lecture.