February 28, 2017

Homework Help: Chemistry

Posted by Graham on Friday, February 1, 2013 at 3:07pm.

I had a 1.094g of mystery Alum. "AB(SO4)c. dH2O" (I assume that that the Alum is KAl(SO4)2-12H2O but we don't know A B c or d, we do know that mass percent of water is 51.9)

After processing the alum (Ba(NO3)2 and HNO3 through a gooch crucible) we end up with 1.739g of BaSO4.

We know that SO4 is 42.15% of BASO4 thus
SO4 has a mass of .7330 g

assuming we captured all of the SO4 of the original Alum.
so if we subtract .7330g from 1.094g = .361 g

therefor .361g is the the AB and the H2O of the Alum

but if water is 51 of the mass of the Alum. I did something wrong any ideas.

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