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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on August 27, 2012, 11:51:28 AM

Title: Problem of the week - 27/08/2012
Post by: Borek on August 27, 2012, 11:51:28 AM
3.0372 g of a mixture containing BaCO3 and BaO2 was dissolved in an excess of a diluted sulfuric acid. Collected precipitate weighed 3.200 g. Filtrate was quantitatively transferred to a 250 mL flask and filled to the mark. Titration of the 25.00 mL of the solution with 0.0200 M permanganate required 9.14 mL of the solution.

What was the composition of the mixture?
Title: Re: Problem of the week - 27/08/2012
Post by: Rutherford on August 28, 2012, 09:33:11 AM
Tried this one. I got that n(H2O2)=4.57*10-3moles. When I assumed that the whole BaO2 reacted (its n is the same as of hydrogen-peroxide) I couldn't get a result that satisfies the two given masses. Then I assumed that not all BaO2 reacted with H2SO4, and I got a system of two equations that couldn't be solved (got negative number). A lower mass of the original mixture could give good answers.
Title: Re: Problem of the week - 27/08/2012
Post by: Borek on August 31, 2012, 07:22:45 AM
A lower mass of the original mixture could give good answers.

3.0372 g of a mixture containing BaCO3 and BaO2

It doesn't say "containing only"...
Title: Re: Problem of the week - 27/08/2012
Post by: Rutherford on August 31, 2012, 07:41:14 AM
Then all BaCO3 and BaO2 precipitated. I got 59.44% of BaCO3 and 25.43% of BaO2.
Title: Re: Problem of the week - 27/08/2012
Post by: Borek on August 31, 2012, 07:52:36 AM
That was probably much easier than the forum upgrade  :P
Title: Re: Problem of the week - 27/08/2012
Post by: Rutherford on August 31, 2012, 08:04:05 AM
Heh, would be harder if not all barium peroxide precipitated and the mixture contained only the carbonate and peroxide (with lower mass), but it was good.
Title: Re: Problem of the week - 27/08/2012
Post by: 123456789 on September 04, 2012, 10:27:14 PM
Sorry this is over, but just wondering is the reaction of KMnO4 and H2O2

This: 2MnO4–(aq) + 3H2O2(aq) → 2MnO2(s) + 2H2O(l) + 3O2(g) + 2OH–(aq)

Or this: 2MnO4- + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O

It is acidic condition, so it should be (2) right? However, some (reputable) sources also claim it to be (1). So which is the correct one? When do you use the 1st eqn then? Thanks!
Title: Re: Problem of the week - 27/08/2012
Post by: Borek on September 05, 2012, 02:41:39 AM
Titration conditions are selected in such a way 2nd equation describes exactly what is happening.

As a rule of thumb permanganate in acidic conditions gets reduced to Mn2+, in neautral to MnO2, in basic to MnO42-. As every rule of thumb it doesn't have to be true always.