Thursday, May 7, 2009

Moles and number of ions?

400. mL of a solution containing 4.80 g/L aluminum sulfate

How do i find the moles. Is this correct.

4.80g Al2(SO4)3 x (1 mole / 342.157g Al2 (SO4)3) = .01403 moles Al2(SO4)3

.0140 moles Al2(SO4)3 x ( 2mole Al / 1 mole Al2(SO4)3) = .02806 mole of Al2

I feel like I am missing something because I do no use the volume of 400 mL. Please help. I need to find the moles of each ion, Al3+ and SO4 2-. If you just do Al, I think I can do the rest.

Moles and number of ions?
.400 L of solution x 4.80 grams/L = 1.92 grams of Aluminum sulfate.

1.92 grams x 1 mole/342.157 g = 0.005611 moles Al2(SO4)3
Reply:ill just agree with you! :)
Reply:You're missing a step.

1. Your first step is OK

2. You only have 0.4L, step 1 assumes one liter. Mulitply by 0.400 to find the moles of aluminum sulfate available.

3. Using the moles available from #2, redo your las equation.

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