Find the mass of AlCl3 that is produced when 25.0 grams of Al2O3 react with excess HCl according to the following equation.

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

Group of answer choices

72.9 g

155 g

65.4 g

16.3 g

32.6 g

Mass = 65.4 g

Explanation:

Given dа ta:

Mass of Al₂O₃ = 25 g

Mass of AlCl₃ = ?

Solution:

Chemical equation:

Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O

Number of moles of Al₂O₃:

Number of moles = mass / molar mass

Number of moles = 25 g / 101.96 g/mol

Number of moles = 0.2452 mol

Now we will compare the moles of Al₂O₃ and AlCl₃.

Al₂O₃ : AlCl₃

1 : 2

0.2452 : 2*0.2452 = 0.4904 mol

Mass of AlCl₃:

Mass = number of moles * molar mass

Mass = 0.4904 mol * 133.34 g/mol

Mass = 65.4 g