Given that there is a 1:1 stoichiometric ratio between aluminum and alum in this reaction, what is the theoretical yield of alum in grams?

Theoretical yield of alum = 24.239 g

Note: No data was provided. The calculations are based on assumed data

Explanation:

Equation of reaction for the production of alum is given below:

2Al + 2KOH + 22H₂0 + 4H₂SO₄ --> 2[KAI(SO₄)₂.12H₂O] + 3H₂

Theoretical is the maximum amount of product that can be theoretically produced by a limiting reactant usually obtained from the calculations using the equation of reaction. The limiting reactant is the reactant that gets used up during the course of the reaction, hence, the reaction stops.

From the equation of reaction, the mole ratio of aluminum to alum is 1:1.

Assuming aluminum is the limiting reactant and the other reactants are in excess in the equation above, and 1.3807 grams of aluminum was used for the reaction, the theoretical yield of alum can be calculated thus;

molar mass of aluminum = 27.0 g/mol; molar mass of alum = 474 g/mol

Number of moles of aluminum present in 1.3807 g = 1.3807 g/ 27 g/mol = 0.051137 moles

Therefore, number of moles of alum produced = 0.051137 moles

mass of o.051137 moles of alum= 0.051137 moles * 474 g/mol =24.239 g

Therefore, theoretical yield of alum = 24.239 g

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