Question

How many grams of AlCl3 can be prepared from 3.5 moles of HCl gas?

Answer 1

Answer:

156 g of AlCl₃ will be produced from 3.5 moles of HCl.

Explanation:

Given data:

Number of moles of HCl = 3.5 mol

Grams of AlCl₃ produced = ?

Solution:

Balanced Chemical equation:

3HCl + Al(OH)₃   →  AlCl₃ + 3H₂O

Now we will compare the moles of AlCl₃  with HCl from balanced chemical equation.

                   HCl          :           AlCl₃

                     3            :              1

                   3.5           :           1/3×3.5 = 1.17 mol

Mass of  AlCl₃ produced:

Mass = number of moles ×molar mass

Mass = 1.17 mol ×   133.341 g/mol

Mass =  156 g

Thus 156 g of AlCl₃ will be produced from 3.5 moles of HCl.