Question

1. Mass of aluminum used - 0.5000 g

2. mass of filter paper and plastic cup used- 4.0088 g

3. mass of alum, filter paper and plastic cup- 10.0024g

4. mass of alum synthesized- 5.9936

5. what is the theoretical yield of alum?

6. what is the percent yield?

Answer 1

The formula of alum is KAl(SO₄)₂^.12H₂O.

2 Al(s) + 2 KOH(aq) + 4 H₂SO₄(aq) + 10 H₂O(l) ⟶ 2 KAl(SO₄)₂∙12 H₂O(s) + 3 H₂(g)

Given that

. Mass of aluminum used - 0.5000 g

Number of mole of Al = amount in g / molar mass

= 0.5000 g /26.981539 g/ mole

= 0.01853 moles Al

Number of moles of Alum= 0.01853 moles Al*2 mole Al /2 mole alum

= 0.01853 moles KAl(SO₄)₂∙12 H₂O(s)

Theoretical yield is calculated as follows:

Amount in g = number of moles * molar mass

= 0.01853 moles KAl(SO₄)₂∙12 H₂O(s)* 474.37 g/mol

=8.7901 g

And here mass of alum synthesized- 5.9936

percent yield = observed mass/ Theoretical yield*100

percent yield of alum =5.9936 g/ 8.7901 g*100

=68.186 %