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.
