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?

Mass = 177.34 g

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

a is the correct answer