[somethingsomethingidk]

Hey I need help with this problem


Al(NO3)3(aq) + 3 KOH(aq) ---> Al(OH)3 (s) + 3 KNO3 (aq)
What mass of aluminum hydroxide could form by the reaction of 40 mL of 0.1 M Al(NO3)3(aq) with an excess of 0.2 M KOH?


I guess what's really confusing me is "with an excess" part. I don't know if an extra step is required there? Or am I looking too deeply into it.
Also, when it says to give mass it wants the answer in grams, correct? But it gives 40 mL in the problem so I assume we deal with molarity so mmol/mL..I guess convert mmol into milligrams? I don't know, I'm so confused with this problem


I tried first converting 0.1 M of Al(NO3)3 into mmols with 40 mL and I got 4 mmol Al(NO3)3 and then converted that into mmols of KOH but that's where I got confused...I wasn't sure if I should use 0.2 or 3 mmol of KOH to convert it.


Please help



*Edit: I got 20.8 mg but I dont think it's right Here's my work:


0.1 mmol Al(NO3)3/mL * 40 mL = 4 mmol Al(NO3)3

4 mmol Al(NO3)3 * 0.2 mmol KOH / 1 mmol Al(NO3)3 = 0.8 mml KOH = 0.0008 mol KOH

0.0008 mol KOH * 1 mol Al(OH)3 / 3 mol KOH = 0.000267 mol Al(OH)3

0.000267 mol Al(OH)3 * 78 grams/mol = 0.0208 grams = 20.8 milligrams

[billnotgatez]

Al(NO₃)₃(aq) + 3 KOH(aq)  Al(OH)₃ (s) + 3 KNO₃ (aq)

@somethingsomethingidk
I rewrote your equations using the editing features in the posting window


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[Borek]

I guess what's really confusing me is "with an excess" part. I don't know if an extra step is required there? Or am I looking too deeply into it.

It just means "there is enough KOH for the reaction to proceed to the end".
 

Also, when it says to give mass it wants the answer in grams, correct? But it gives 40 mL in the problem so I assume we deal with molarity so mmol/mL..I guess convert mmol into milligrams?

Molarity is mol/L, but you are on the right track.

I tried first converting 0.1 M of Al(NO3)3 into mmols with 40 mL and I got 4 mmol Al(NO3)3

That was OK. Now, don't look at KOH at all - it doesn't matter (you know there is enough, and that's the only important thing). How many moles of Al(OH)₃ will be produced?