When a mixture of aluminum powder and iron(III) oxide is ignited, it produces molten iron and aluminum oxide. ?

When a mixture of aluminum powder and iron(III) oxide is ignited, it produces molten iron and aluminum oxide. In an experiment, 5.40 g of aluminum was mixed with 18.90 g of iron(III) oxide. At the end of the reaction, the mixture contained 11.17 g of iron, 9.60 g of aluminum oxide, and an undetermined amount of unreacted iron(III) oxide. No aluminum was left. What is the mass of the iron(III) oxide?





anybody know?When a mixture of aluminum powder and iron(III) oxide is ignited, it produces molten iron and aluminum oxide. ?
First you must write a balanced equation for the reaction and indicate the amounts given:





2 Al + Fe2O3 --%26gt; Al2O3 + 2 Fe


5.40 g 18.90 g 9.60 g 11.17 g





We also know that some iron (III) oxide what left unreacted, indicating that it was present in excess.





According to the law of conservation of mass, the total mass of reactants (aluminum + iron (III) oxide) is equal to the total mass of products (aluminum oxide + iron). Since we know the amounts of the products formed (9.60 g Al2O3 + 11.17 g Fe = 20.77 g total products), we can assume that 20.77 g of reactants have been used up in this reaction as well. We know that all the aluminum (5.40 g) reacted, so by subtracting its mass from the total (20.77g - 5.40 g Al = 15.37 g Fe2O3) we find the amount of iron (III) oxide that actually reacted with aluminum. Subtracting this amount from the available iron (III) oxide (18.90 g - 15.37 g = 5.53 g left over) we end up with the amount of Fe2O3 that was left.