Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.0 g of HCl according to the following equation. Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(aq)

The mass of aluminium chloride produced is 12.17 grams

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of aluminium oxide = 10.0 g

Molar mass of aluminium oxide = 102 g/mol

Putting values in equation 1, we get:

Given mass of HCl = 10 g

Molar mass of HCl = 36.5 g/mol

Putting values in equation 1, we get:

The given chemical equation follows:

By Stoichiometry of the reaction:

6 moles of HCl reacts with 1 mole of aluminium oxide

So, 0.274 moles of HCl will react with = of aluminium oxide

As, given amount of aluminium oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, HCl is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

6 moles of HCl produces 2 moles of aluminium chloride

So, 0.274 moles of HCl will produce = of aluminium chloride

Now, calculating the mass of aluminium chloride from equation 1, we get:

Molar mass of aluminium chloride = 133.34 g/mol

Moles of aluminium chloride = 0.0913 moles

Putting values in equation 1, we get:

Hence, the mass of aluminium chloride produced is 12.17 grams