A solution contains Al3 and Co2 . The addition of 0.3868 L of 1.707 M NaOH results in the complete precipitation of the ions as Al(OH)3 and Co(OH)2 . The total mass of the precipitate is 23.36 g . Find the masses of Al3 and Co2 in the solution.

See Explanation

Explanation:

Number of moles of NaOH = 0.3868 L × 1.707 M = 0.660 moles

Reaction of Al^3+ and OH^-

Al^3+(aq) + 3OH^-(aq) > Al(OH)3(s)

If 1 mole of Al3+ reacts with 3 moles of OH^-

x moles of Al^3+ reacts with 0.660 moles of OH^-

x = 1 × 0.660/3

x= 0.22 moles

Mass of Al^3+ = 0.22 moles × 27 g/mol = 5.94 g of Al^3+

Co^2+(aq) + 2OH^-(aq) > Co(OH)2(s)

1 mole of Co^+ reacts with 2 moles of OH^-

x moles of Co^2+ reacts with 0.660 moles of OH^-

x = 1 × 0.660/2 = 0.33 moles

Mass of Co^2+ = 0.33 moles × 59 g/mol = 19.47 g

i dont know

explanation: