The reaction between aluminum and iron(iii) oxide can generate temperatures approaching 3000°c and is used in welding metals: 2al + fe2o3 → al2o3 +2fe in one process, 149 g of al are reacted with 601 g of fe2o3. calculate the mass (in grams) of al2o3 formed, and determine the amount of excess reagent left at the end of the reaction.

First, we need to find the limiting reactant. 

149 g Al x 1 mole/27g Al = 5.52 mol Al 

601 g Fe2O3 x 1 mole/159.6g Fe2O3 = 3.77 mol Fe2O3 

Al is the limiting reactant since 3.77 mol Fe2O3 would require 3.77*2 = 7.54 mol Al but the given is less.

mol Al2O3 = 5.52 mol Al  * 1 mol Al2O3 / 2 mol Al = 2.76 mol

mass Al2O3 = 2.76 mol x 102g/1 mole AlO3 = 281.52 g Al2O3 will be formed. 

Since 3.77 mole of Fe2O3 is present, but 5.52 mole Al is the limiting reactant, then only 5.52/2 mole Fe2O3 can be used. This leaves an excess of:

Fe2O3 excess = 3.77 – 5.52/2 = 1.01 

1.01 mole Fe2O3 x 103.8g/1 mole Fe2O3 = 104.84 g Fe2O3