Write balanced total and net ionic equations for its reaction with aqueous NaOH?

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.


Total equation:
1 Al2(SO4)3 (aq) +6 NaOH (aq) ----> blank (aq) + blank (s)

net ionic:
blank (aq) + blank (aq) ---> blank (s)



B) What mass (g) of precipitate forms when 137.5 mL of 0.636 M NaOH is added to 517 mL of a solution that contains 18.0 g aluminum sulfate per liter?



If anyone has any ideas please help with what you can. Thank you sooo much.

Answer 1

First question:


The balanced equation looks like this:

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

net ionic equation looks like this:

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

The white precipitate is the Aluminium Hydroxide.

Second part (B):

Oops! Next friend has completed the answer. He is right. Just a correction to avoid confussion: he wrote somewhere Na2(SO4)3 instead of Al2(SO4)3. Thanks anyway to let me in.

Answer 2

Total equation:

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

net ionic equation:

Al3+ + 3OH- -> Al(OH)3 (<- this is the precipitate)

B. The NaOH solution contains:

n = C*V = 0.636*0.1375 = 0.8745 mol NaOH

The molar weight of Al2(SO4) equals to Mr = 342 g/mol, so 18 g of the salt are:

n = m/Mr = 18/342 = 0.053 mol Al2(SO4)3.

Now, from the chemical equation 1 mol of Al2(SO4)3 reacts with 6 mol of NaOH, so the 0.053 mol of Na2(SO4)3 reacts with 0.053*6 = 0.316 mol of NaOH. But we have 0.8745 mol of NaOH. So NaOH is in excess and the Al2(SO4)3 reacts fully.

1 mol of Al2(SO4)3 gives 2 mol of Al(OH)3, so 0.053 mol of Al2(SO4)3 give 2*0.053 = 0.106 mol of Al(OH)3. The molar mass of Al(OH)3 is 78 so:

mass of Al(OH)3 (precipitate) = n*Mr = 0.106*78 = 8.21 g.