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!
chemistry problem
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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.
https://en.wikipedia.org/wiki/Aluminium_sulfide
Fe2(CrO4)3 see
https://en.wikipedia.org/wiki/Iron(III)_chromate
both not existing in solution.