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)

net ionic equation contains only reacting ions

from the reaciton of

ALuminium sulfate + base

Al2(SO4)2(aq) + NaOH(aq) = Al(OH)3(s) + Na2SO4(aq)

the ions reaciting :

Al3+(Aq) + OH-

net ionic:

Al3+(aq) + 3OH-(aq) --> Al(OH)3(s)

b)

find moles of Al3+

mol of Al2(SO4)2 ---> mass/MW = 19.7/342,1509 = 0.057576 mol of  Al2(SO4)2

mol of Al =  Al2(SO4)2 *2 =0.057576 *2 = 0.115152 mol of Al

now...

this is in 1 liter, but we have 517 mL so

fraction of Al --> 517/1000*0.057576 = 0.02977 mol of Al

now...

mol of NaOH = MV = (0.749)(168*10^-3) = 0.125832 mol of NaOH

OH- mol = 0.125832

ratio is

1 mol of Al = 3 mol of OH-

0.02977 mol of Al will required --> 3x0.02977 = 0.08931 mol of OH-

now... Al is limiting

so...

mol of Al(OH) expected to form

0.02977 mol ofAl --> 0.02977 mol of AL(OH)3

mass = mol*MW = 0.02977 *(78) = 2.32206 g of Al(OH)3 expected