Answer:
The answer to your question is: 70.8%
Explanation:
Data
Al₂O₃ = 60 g
C = 30 g
CO = gas
Al = 22.5 g
MW Al₂O₃ = 102 g
MW C = 12 g
MW Al = 54 g
Reaction
Al₂O₃ + 3C ⇒ 3 CO + 2 Al
Limiting reactant
102 g of Al₂O₃ -------------- 54 g Al
60 g -------------- x
x = 31.8 g
36 g of C ------------------ 54 g of Al
30 g of C ------------------ x
x = 45 g of Al
Limiting reactant = Al₂O₃
Percent yield = 
Percent yield = 70.75 %
Well ask yourself why don't we count it in moles and you should get your answer.
I would answer this but i have to go to bed i have a reading test tommorow
Most elements are metals. A.
N₂ + 3H₂ ⇒ 2NH₃
1mol : 2mol
3,72mol : 7,44mol
n = 7,44mol
M = 17g/mol
m = n * M = 7,44mol * 17g/mol = 126,48g