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iknowashtron
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Joined: 07 Oct 2017
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PostPosted: Sat Oct 07, 2017 3:35 am    Post subject: chemistry problem Reply with quoteFind all posts by iknowashtron

If i have 100mL of the solution iron(III) chromate having a concentration of 0,620 mol/L.
All Fe3+ ions are precipitated by adding to this solution a certain volume of aluminum sulfide solution of 1,00mol/L.
Considering that the reaction is complete, that there is no excess and that the volumes can be added, how do I calculate the concentration of Al3+ ions after precipitation.
I think I completed and balanced the equation correctly, but I’m not sure:
2Fe2(CrO4)3 + Al2S3 = Fe4S3 + 2Al(CrO4)3
Then I think I'm supposed to use c = n/v somehow to calculate the concentration?
Could someone walk me through what i'm supposed to do? Thank you!
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ChenBeier
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Location: Berlin, Germany
PostPosted: Sat Oct 07, 2017 6:17 am    Post subject: Reply with quoteFind all posts by ChenBeier

Al2S3 see

https://en.wikipedia.org/wiki/Aluminium_sulfide

Fe2(CrO4)3 see

https://en.wikipedia.org/wiki/Iron(III)_chromate

both not existing in solution.
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