Use the balanced equation below: How many grams of molten iron can be produced if you have 100 grams of Iron Oxide and 200 grams of Aluminum?

2 Al(s) + Fe2O3(s)  Al2O3(s) + 2 Fe (l)

You need to find the amount of moles of iron (III) oxide and aluminium you have. Then you look at your equation, and deduce which one of the two is the limiting reagent (which means not in excess).

If you have excess, the excess will not react, hence the need for you to find the limiting reagent.

The number of moles of iron (III) oxide is given by the total mass (100 g) divided by the molecular mass of iron (III) oxide.
The number of moles of aluminium is given by the total (200 g) divided by the molecular mass of aluminium.

When you found the number of moles of the limiting reagent, find the number of moles of Fe that is produced. From that find its mass.

Post your answer! :)