Net Ionic equation

[pmandola]

Please help.
My Chemistry homework states:
[i]Write balanced total and net ionic equations for Aluminum sulfate's reaction with aqueous Sodium hydroxide. [/i]
My balanced equation is:
Al₂(SO₄)₃ + 6 NaOH = 2 Al(OH)₃ + 3 NaSO₄
I believe my net ionic equation would be:
2Al₃+ + 6OH- = 2Al(OH)₃
The homework program says my coefficents are incorrect.

To end this same problem:
What mass of precipitate forms when 135.0mL of 0.625M NaOH is added to 517mL of a solution that contains 16.7g aluminum sulfate per liter?

I've tried this several times but I'm obviously not laying out the conversions correctly.

[peter]

Coefficients in a chyemical equation always must be the lowest possible integers, there equation:

Code: Select all

2Al{3+} + 6OH{-} = 2Al(OH)3 

must be rewritten as:

Code: Select all

Al{3+} + 3OH{-} = Al(OH)3 

[pmandola]

Thanks! I knew I was missing something obvious.

What about the remainder of the question:
To end this same problem:
What mass of precipitate forms when 135.0mL of 0.625M NaOH is added to 517mL of a solution that contains 16.7g aluminum sulfate per liter?