Calculating Unreacted Al in Thermite Reaction

[chemister]

I have a question regarding the thermite reaction.

The chemical equation for it looks like:

$$Fe_{2}O_{3}+2Al----2Fe+Al_{2}O_{3}$$

If 1 mole of Al is reacted with 10.0g of $$Fe_{2}O_{3}$$, how many grams of unreacted Al would remain?

Would you just convert the 10.0g to moles which would be 0.0626mol and then subtract from 1? Or do you have to us some kind of ratio too?

Thanks!

 

[poolwin2001]

Basically 1mole$$Fe_{2}O_{3}$$reacts with two moles of Al.

 

[chemister]

Huh? I don't understand?

 

[mrjeffy321]

for every 1 mol of Iron Oxide (Fe2O3) you have, you will need 2 mols of Aluminum (Al) to react it fully.

if you know how much of both you have, then this will be easy.
it is pretty obvious, if not almost blatantly stated in the question that aluminum is in excess and Iron Oxide is your limiting reagent.
so convert the gram of Iron Oxide into mols by dividing the mass by the molar weight. then you know that you will need twice as many mols of aluminum as you have Iron Oxide to react it, so double it to get the number of mols of aluminum.
You know how many mols of aluminum you started out with (1), so then subtract the used mols of aluminum from how many you have, and that is how many are left over. after you get that, just convert that into grams by mltiplying by the molar weight of aluminum.