Question

. In a precipitation reaction between silver nitrate and aluminum chloride, 25.5 mL of an AlCl3 solution was required to precipitate 4.90 g of silver chloride. What was the molarity of the AlCl3 solution? You will need to write and balance the equation for the precipitation reaction first.

Answer 1

3AgNO3 + AlCl3 → Al(NO3)3 + 3AgCl
3 mol AgNO3 react with 1 mol AlCl3 to produce 3 mol AgCl

Mol AgCl in 4.90g AgCl
Molar mass AgCl = 143.3g/mol
mol AgCl = 4.90/ 143.3 = 0.0342 mol AgCl
3 mol AgCl were produced from 1 mol AlCl3
0.0342 mol AgCl were produced from 0.0342/3 = 0.0114 mol AlCl3

Now you know that 25.5mL of AlCl3 solution contain 0.0114 mol AlCl3
1000mL of AlCl3 solution contain : 1000/25.5 * 0.0114 = 0.447 mol AlCl3
Molarity of AlCl3 solution = 0.447 M