Question

8Al + 3Fe3O4 = 4Al2O3 + 9Fe. A) how many grams of iron are produced by the reaction of 225.0 grams of Al and 225.0 grams of Fe3O4? B) How many grams of Al2O3 are also produced in the reaction of (A)? C) which reactant is in excess and how many grams remain after the reaction?

Answer 1

For these kinds of questions always consider moles rather than masses, because chemical reactions work with the NUMBER rather than the MASS of the compounds involved.

A) Convert the masses of Al and Fe3O4 into moles (divide by molar mass). You will need to determine the limiting reactant here since you are given starting amounts of both reactants (divide moles by the corresponding coefficient in the chemical reaction - the lower number is limiting). Use the moles of the limiting reactant to find the moles of Fe produced (using stoich ratios), then convert that to grams (multiply by the molar mass).

B) Set up the same stoich ratio equation as in A), but use the coefficient of Al2O3 rather than Fe. Then multiply by the appropriate molar mass.

C) Set up another stoich ratio using either of the products or the limiting reactant to find the moles of the excess reactant consumed. Convert this value into mass as usual, and subtract that mass from the total starting mass of that reactant to find the excess.

Let me know if you have any questions!

Answer 2

A) 225.0g Al / 26.98g Al * 9mol Fe / 8molAl * 55.85g Fe = 523.9g Fe
225.0g Fe3O4 / 231.6g Fe3O4 * 9 mol Fe / 3mol Fe3O4 * 55.85g Fe = 162.8g Fe


B) 225.0 g Al / 26.98g Al * 4 mol Al2O3 / 8 mol Al * 128.9g AL2O3 = 537.5g Al2O3
225.0g Fe3O4 / 231.6g Fe3O4 * 4 mol AL2O3 / 3 mol Fe3O4 * 128.9g Al2O3=166.9g Al2O3

C) ?????