Answer:
The theoretical yield of alum is 19.873 grams and percent yield for this alum synthesis is 45.33%.
Explanation:
Mass of bottle = 10.221 g
Mass of bottle with aluminium pieces = 11.353 g
Mass of aluminium = 11.353 g - 10.221 g = 1.132 g
Mass of alum and bottle = 19.230 g
Mass of alum = 19.230 g - 10.221 g = 9.009 g
Experimental yield of alum = 9.009 g
Theoretical yield of alum:

Moles of aluminium = 
According to reaction, 2 moles of aluminum gives 2 moles of alum.
Then 0.041926 mol aluminium will give :
of alum.
Mass of 0.041926 moles of alum:
0.041926 mol × 474 g/mol= 19.873 g
Theoretical yield of alum = 19.873 g
Percentage yield:

Percentage yield of the alum:

The theoretical yield of alum is 19.873 grams and percent yield for this alum synthesis is 45.33%.