VCE CHEM in uni problem- limiting reagent

[Starlight]

The question is:

a reaction mixture contains 400g 02 and 270g Al. The aluminium reacts with the oxygen (02) to produce Aluminium oxide A203.

Balanced equation: 4Al + 302 ---> 2Al203

No moles each reactant: n(Al)= 10 moles, n(02)= 12.5

This part is what i'm stuck on: Explain which one is the limiting reagent for this reaction.

The solution says: 10.0 moles of Al will react with 7.5 moles of O2, therefore Al is the limiting reagent. What i'm wondering is where they get the 7.5 moles of 02 from?

[Somye]

using stoich ratios, your n(O2) reqd = 3/4 n(Al)
which means n(O2) reqd =  .75(10)
=7.5mol
therefore, Al is your limiting reagent

[golden]

The question is:

a reaction mixture contains 400g 02 and 270g Al. The aluminium reacts with the oxygen (02) to produce Aluminium oxide A203.

Balanced equation: 4Al + 302 ---> 2Al203

No moles each reactant: n(Al)= 10 moles, n(02)= 12.5

This part is what i'm stuck on: Explain which one is the limiting reagent for this reaction.

The solution says: 10.0 moles of Al will react with 7.5 moles of O2, therefore Al is the limiting reagent. What i'm wondering is where they get the 7.5 moles of 02 from?

According to the equation 4 moles of Al will react with 3 moles of oxygen. If you have 10 moles of Al then because of the co-oefficients then you will get 3/4 x 10 = 7.5 moles of oxygen.

[Starlight]

Thanks guys! It's always the basics that stuffs me up