Find the mass of AlCl3 that is produced when 3.58 moles of Al2O3 react with excess HCl according to the following equation.

Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l)

Moles of Al2O3, m = 3.58 moles.

We need to find the mass of AlCl3 produced.

Solution :

1 mole of Al2O3 produce 2 moles of AlCl3.

So, 3.58 moles of AlCl3 produce 7.16 moles of AlCl3.

Mass of AlCl3 produced is :

m = moles × molecular mass of AlCl3

m = 7.16 × 133.34 g

m = 954.714 g

Therefore, mass of AlCl3 produced is 954.714 g.

Hence, this is the required solution.

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