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

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:

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

Amount of Alum produced = 17.34 grams

Theoretical yield of alum = 17.34 grams.