Answer:
The percent yield of the reaction is 82%
Explanation:
First step: make the chemist equation.
2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)
As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3
Second step: Find out the moles in the reactant.
Molar weight Fe2O3: 159.7 g/m
Mass / Molar weight = moles
50 g /159.7 g/m = 0.313 moles
Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.
1 mol Fe2O3 ____ 2Fe
0.313 mol Fe2O3 ____ 0.626 moles
Molar weight Fe = 55.85 g/m
Moles . molar weight = mass
55.85g/m . 0.626m = 34.9 grams
This will be the 100% yield of the reaction but we only made 28.65 g
34.9 g ____ 100%
28.65 g ____ 82.09 %
