Answer:
92.75%
Explanation:
The overall chemical equation for the reaction in the preparation of alum from the aluminium can be expressed as:
[tex]\mathtt{2Al + 2KOH + 4H_2SO_4 +2H_2O \to 2KAl(SO_4)_2 2H_2O +3H_2}[/tex]
From above; we will see that 2 moles of Aluminium react with sulphuric acid and water to produce 2 moles o aluminium alum.
Therefore, the theoretical yield can be determined as:
[tex]=0.4395 \ g Al \times \dfrac{1 \ mol \ Al}{27 \ g Al }\times \dfrac{2 \ mol \ KAl(SO_4)_2}{2 \ mol \ Al}\times \dfrac{294.23 \ KAl(SO_4)_2}{1 \ mol \ KAl(SO_4)_2}[/tex]
= 4.789g of [tex]KAl(SO_4)_2[/tex]
To find the percent yield, we need to divide the actual yield by the theoretical yield and then multiply it with 100.
∴
percent yield = ( mass of alum(g)/theoretical yield(g) ) × 100
percent yield = ( 4.789g / 5.1629g ) × 100%
percent yield = 0.9275 × 100%
percent yield = 92.75%
Thus, the percent yield of the experiment 92.75%
