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)₄^2-. 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

Molecular balanced equation

Aluminum sulfate + base

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

balanced

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

net ionic must contain only reacting species

then--> 2Al(OH)3(s) product, contains ions Al+3 and OH-

so

ne tionic:

2Al3+(aq) + 6OH-(aq) = 2Al(OH)3(s)

b)

mass of precipitate if:

mmol of base = MV = 0.749*168 = 125.832 mmol of OH-

find mass of Al2(SO4)3;

517*19.7/1000 = 10.1849 g of Al2(SO4)3;

change mol

mol = mass/MW = (10.1849)/(342,150) = 0.0000297 mol --> 0.0297 mmol

then

Al+3 is limiting

2 mol of Al+3 = 1 mol of Al2(SO4)3

0.0297 mmol --> 1/2*0.0297 = 0.01485 mmol of Al(OH)3

mass = mmol*MW = 0.01485 *78.003558 = 1.158 mg