Question

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

Answer 1

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

Step-by-step 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.