Fe2O3(s) + 2 Al(s)→ 2 Fe(l) + Al2O3(s)?

What masses of iron(III) oxide and aluminum must be used to produce 16.0 g iron? What is the maximum mass of aluminum oxide that could be produced?

Answer:

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16g Fe x (1 mol/55.85 g Fe) x (1 mol Fe2O3/2 mol Fe) x (159.7 g Fe2O3/1 mol Fe2O3) 22.88 g Fe2O3 (This gives the mass of Fe2O3) 16 g Fe x (1 mol Fe/55.85 g Fe) x (2 mol Al/2 mol Fe) x (26.98 g Al/1 mol Al) 7.73 g Al (Mass of Al) Now choose either Fe2O3, Al, or Fe to find Al2O3(Its best to choose the given amount to ensure you are entirely accurate to avoid any errors due to rounding on your other compounds.) 16 g Fe x (1 mol Fe/55.85 g Fe) x (1 mol Al2O3/2 Fe) x (101.96 g Al2O3/1 mol Al2O3) 14.6 g Al2O3 These all of course assume 100% yieldIf it specifies that only 77% yield occured in the reaction for instance, multiply your you Fe2O3 and Al by 1/.77 to find their original amounts.