[chay722]

If 37.8 g of aluminum is reacted with 91.3 g of iron (III) oxide, what mass of iron will be produced? State the given and the unknown.

I set up a table to do this. The balanced equation I have is: 2Al+Fe2O3 yields Al2O3+2Fe. I converted the given grams into the proper moles and thus I have 1.4 moles under the 2Al and .57 moles under the Fe2O3. In any event, I came up with an answer of 63.67 grams. Is this correct? Also, I'm not sure what they mean when they ask, "state the given and the unknown." Thanks for any help.

[Mitch]

The 37.8 g of aluminum and 91.3 g of iron (III) oxide are givens

[chay722]

Thank you. So the unknown would just be the amount they do not give us, correct?

[chay722]

In any event, I came up with an answer of 63.67 grams. Is this correct?

[Yggdrasil]

Looks right to me.

[AWK]

Just look at balanced reaction. Fe2O3 is a limited reagent. 0.57 Mole of Fe2O3 needs 1.14 mole of Al2O3 (1.40-1.14=0.26 mole Al - extent)  and  1.14 mole Fe will be formed(mass of Fe is 1.14x56).

Of couse, your result is OK