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

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

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

2Al+3+ 6OH- --------->2Al(OH)3 net ionic equation

the molecular reaction suggests 1 mole of Al2(SO4)3 reacts with 6 mole of NaOH to precipitate 2 mole of Al(OH)3

molar ratio of Al2(SO4)3: NaOH= 1: 6

mass of Al2(SO4)3= mass/liter* volume in L= 19.7 g/L * 517/1000 L = 10.2 gm

moles of Al2(SO4)3= mass/molar mass =342, moles of Al2(SO4)3= 10.2/342= 0.03

moles of NaOH= molarity* Volume in L= 0.749*168/1000 =0.1258

actual molar ratio of Al2(SO4)3: NaOH= 0.03:0.1258= 1:4.2

so limiting reagent is NaOH since required ratio is 6 where as supplied ratio is 4

hecne 6 mole of NaOH gives 2 moles of Al(OH)3

0.1258 moles gives 0.1258*2/6 moles of Al(OH)3=0.042, molar mass of Al(OH)3= 78 g/mole

mass of Al(OH)3 formed= moles* molar mass= 0.042*78= 3.3 gm