Question

Al2O3 + 6 HCl ↔ 2 AlCl3 + 3 H2O If you start with 67.7 grams Al2O3 and 196.8 grams HCl, how many grams of AlCl3 can be produced by the reaction?

Answer 1

Mass = 177.34 g

Step-by-step explanation:

Given data:

Mass of Al₂O₃ = 67.7 g

Mass of HCl = 196.8 g

Mass of AlCl₃ = ?

Solution:

Chemical reaction:

Al₂O₃ +6 HCl → 2AlCl₃ + 3H₂O

Number of moles of Al₂O₃:

Number of moles = mass/ molar mass

Number of moles = 67.7 g / 101.96 g/mol

Number of moles = 0.664 mol

Number of moles of HCl:

Number of moles = mass/ molar mass

Number of moles = 196.8 g / 36.5 g/mol

Number of moles = 5.4 mol

Now we will compare the moles of AlCl₃ with HCl and Al₂O₃.

Al₂O₃ : AlCl₃

1 : 2

0.664 : 2×0.664 = 1.33 mol

HCl : AlCl₃

6 : 2

5.4 : 2/6×5.4 = 1.8 mol

The number of moles produced by Al₂O₃ are less so it will limiting reactant.

Mass of AlCl₃:

Mass = number of moles × molar mass

Mass = 1.33 mol × 133.34 g/mol

Mass = 177.34 g