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:
![2Al(s) +2 KOH(aq) +4H_2SO_4(aq)+10 H_2O(l) \rightarrow 2 KAl(SO_4)_2.12 H_2O(s)+3H_2(g)](https://tex.z-dn.net/?f=2Al%28s%29%20%2B2%20KOH%28aq%29%20%2B4H_2SO_4%28aq%29%2B10%20H_2O%28l%29%20%5Crightarrow%202%20KAl%28SO_4%29_2.12%20H_2O%28s%29%2B3H_2%28g%29)
Moles of aluminium = ![\frac{1.132 g}{27 g/mol}=0.041926 mol](https://tex.z-dn.net/?f=%5Cfrac%7B1.132%20g%7D%7B27%20g%2Fmol%7D%3D0.041926%20mol)
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:
![\% Yield =\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100](https://tex.z-dn.net/?f=%5C%25%20Yield%20%3D%5Cfrac%7B%5Ctext%7BExperimental%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100)
Percentage yield of the alum:
![\% Yield =\frac{ 9.009 g}{19.873 g}\times 100=45.33\%](https://tex.z-dn.net/?f=%5C%25%20Yield%20%3D%5Cfrac%7B%209.009%20g%7D%7B19.873%20g%7D%5Ctimes%20100%3D45.33%5C%25)
The theoretical yield of alum is 19.873 grams and percent yield for this alum synthesis is 45.33%.