Question

Aluminum sulfate, known as cake alum, has a remarkably wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate.

(a) Write balanced total and net ionic equations for its reaction with aqueous NaOH. (Type your answer using the format (NH4)2CO3 for (NH4)2CO3, [NH4]+ for NH4+, and [Ni(CN)4]2- for Ni(CN)42-. Use the lowest possible coefficients. Type the cation before the anion.)

(b) What mass of precipitate forms when 168.0 mL of 0.749 M NaOH is added to 517 mL of a solution that contains 19.7 g aluminum sulfate per liter?

Answer 1

a)

molecular equation :

Al2(SO4)3 (aq) + 6 NaOH (aq) -------------------> 2 Al(OH)3 (s) + 3 Na2SO4 (aq)

total ionic equation

2Al+3(aq) + 3SO4-2 (aq) + 6 Na+ (aq) + 6OH- (aq) -------------------> 2 Al(OH)3 (s) + 6Na+ (aq) + 3SO4-2 (aq)

net ionic equation :

2 Al⁺³(aq) + 6OH^- (aq) -------------------> 2 Al(OH)₃ (s)

b)

moles of NaOH = 168 x 0.749 / 1000 = 0.126

moles of Al2(SO4)3 = 19.7 / 342.1 = 0.0576

Al2(SO4)3 (aq) + 6 NaOH (aq) -------------------> 2 Al(OH)3 (s) + 3 Na2SO4 (aq)

1                                  6

0.0576                       0.126

limiting reagent is NaOH :

moles of Al(OH)3 = 0.126 / 6

                             = 0.021

mass of Al(OH)3 = 0.021 x 78.0

                           = 1.64 g

mass of precipitate forms = 1.64 g