Question

Calculate the theoretical yield of alum expected from 0.9875 g of aluminum foil. assume the aluminum is the limiting reactant.

Answer 1

Answer: 17.34 grams of alum will be produced if 0.9875 g of Aluminium foil was used.

Explanation: Reaction to form alum from Aluminium is given as:

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

We are given Aluminium to be the limiting reactant, so the formation of alum will be dependent on Aluminium because it limits the formation of product.

By stoichiometry,

2 moles of Al is producing 2 moles of Alum

Mass of 2 moles of Aluminium = (2 × 27)g/mol = 54 g/mol

Mass of 2 moles of alum = (2 × 474)g/mol = 948 g/mol

54 g/mol of aluminium will produce 948 g/mol of alum, so

$\text{0.9875 grams of aluminium will produce}=\frac{948g/mol}{54g/mol}\times 0.9875g$

Amount of Alum produced = 17.34 grams

Theoretical yield of alum = 17.34 grams.

Answer 2

The theoretical yield of alum expected from 0.9875 g of aluminum foil is 17.36 g.

Let's consider the overall equation for the obtaining of alum from aluminum

2 Al(s) + 2 KOH(aq) + 2 H₂O(l) + 4 H₂SO₄(aq) ⇒ 2 KAl(SO₄)₂(s).12H₂O(l) + 3 H₂(g)

We can calculate the theoretical yield of alum from 0.9875 g of Al considering the following relationships:

• The molar mass of Al is 26.98 g/mol.

• The molar ratio of Al to alum is 2:2.

• The molar mass of alum is 474.39 g/mol.

$0.9875gAl \times \frac{1molAl}{26.98gAl} \times \frac{2molAlum}{2molAl} \times \frac{474.39gAlum}{1molAlum} = 17.36gAlum$

The theoretical yield of alum expected from 0.9875 g of aluminum foil is 17.36 g.

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