Question

Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be synthesized from aluminum metal, sulfuric acid, water, and potassium hydroxide, as seen in the equation. 2 Al(s) 2 KOH(aq) 4 H2SO4(aq) 10 H2O(l) ⟶ 2 KAl(SO4)2∙12 H2O(s) 3 H2(g) alum Using the data, determine the theoretical and percent yield for this alum synthesis. Note that aluminum is the limiting reactant. Description Mass (g) bottle mass 10.221 bottle mass with aluminum pieces 11.353 final product and bottle mass 19.230 What is the theoretical yield of alum

Answer 1

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:

$2Al(s) +2 KOH(aq) +4H_2SO_4(aq)+10 H_2O(l) \rightarrow 2 KAl(SO_4)_2.12 H_2O(s)+3H_2(g)$

Moles of aluminium = $\frac{1.132 g}{27 g/mol}=0.041926 mol$

According to reaction, 2 moles of aluminum gives 2 moles of alum.

Then 0.041926 mol aluminium will give :

$\frac{2}{2}\times 0.041926 mol=0.041926 mol$ 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:

$\% Yield =\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the alum:

$\% Yield =\frac{ 9.009 g}{19.873 g}\times 100=45.33\%$

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