Question

g A 0.4395 g sample of aluminum reacts according to our experiment to produce alum. 5.1629 g of dried alum crystals are recovered. What is the percent yield of the experiment?

Answer 1

Answer:

92.75%

Explanation:

The overall chemical equation for the reaction in the preparation of alum from the aluminium can be expressed as:

$\mathtt{2Al + 2KOH + 4H_2SO_4 +2H_2O \to 2KAl(SO_4)_2 2H_2O +3H_2}$

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:

$=0.4395 \ g Al \times \dfrac{1 \ mol \ Al}{27 \ g Al }\times \dfrac{2 \ mol \ KAl(SO_4)_2}{2 \ mol \ Al}\times \dfrac{294.23 \ KAl(SO_4)_2}{1 \ mol \ KAl(SO_4)_2}$

= 4.789g of $KAl(SO_4)_2$

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%