Answer to Question #71269 in General Chemistry for Daniel

Question

What is the maximum number of grams of Al2O3 that can be made from the reaction of 48.0 g of Al with 225 g of Fe2O3?

Answer 1

2Al + Fe2O3 = 2Fe + Al2O3
n(Al) = 48 / 27 = 1.78 (mol)
n(Fe2O3) = 225 / (56*2 + 16*3) = 1.40 (mol) – limiting reagent
n(Al2O3) = 1.40 mol
m(Al2O3) = n * M = 1.40 * (27*2 + 16*3) = 142.8 (g)