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Topic: Another Stoichiometry Problem  (Read 3928 times)

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Offline positiveion

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Another Stoichiometry Problem
« on: October 25, 2009, 02:14:53 AM »
14.48 g of a metal sulfate with the formula M2SO4 were dissolved in water. Excess barium nitrate solution was added in order to precipitate all the sulfate ions in the form of barium sulfate. 9.336 g of precipitate was obtained.

a). calculate the amount of barium sulfate BaSO4 precipitated.


I have no idea how to do this problem.

Offline ShadowSpirit

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Re: Another Stoichiometry Problem
« Reply #1 on: October 25, 2009, 03:19:30 AM »
But you already said 9.336g of precipitate was obtained? So that would be the amount of BaSO4?

If you need the mole, you could use mass/molar mass~

Offline positiveion

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Re: Another Stoichiometry Problem
« Reply #2 on: October 25, 2009, 03:23:44 AM »
I have no idea. I just typed out the question.

The answer says 4.00 * 10^-2

Offline ShadowSpirit

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Re: Another Stoichiometry Problem
« Reply #3 on: October 25, 2009, 03:34:05 AM »
That would be the number of moles of BaSO4, which you can obtain by mass/molar mass:

9.336/ (137.3+32.1+16x4) = 0.0400

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