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Author Topic: Problem of the week - 10/12/2012  (Read 3679 times)

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Borek

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Problem of the week - 10/12/2012
« on: December 09, 2012, 11:34:17 PM »

Mixture of CaO and CaCO3 was reacted stoichiometrically with a solution of hydrochloric acid, producing 8.04 L of CO2 (measured at 290 K and 120 kPa). Mass of the dry CaCl2 obtained was 50.61% higher than the mass of the original mixture. How much water was present in the 1M HCl solution used for the reaction, if its density was 1.02 g/mL?
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XGen

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Re: Problem of the week - 10/12/2012
« Reply #1 on: December 10, 2012, 02:37:29 PM »

1.95 kg?
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Bublik

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Re: Problem of the week - 10/12/2012
« Reply #2 on: December 10, 2012, 04:56:06 PM »

I get 1.91L :S
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curiouscat

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Re: Problem of the week - 10/12/2012
« Reply #3 on: December 11, 2012, 07:35:34 PM »

1965 gm?

PS. Why do we not consider this reaction:

CaO + H2O  :rarrow: Ca(OH)2

Does it not occur? Or will HCl ultimately convert Ca(OH)2 to CaCl2?

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Borek

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Re: Problem of the week - 10/12/2012
« Reply #4 on: December 11, 2012, 10:21:27 PM »

1965 gm?

This is closest to the number I was looking for. I got 1967 g. Could be the difference is in the molar masses, I am using EBAS which means I am using exact numbers, not approximate ones.

Quote
PS. Why do we not consider this reaction:

CaO + H2O  :rarrow: Ca(OH)2

Does it not occur? Or will HCl ultimately convert Ca(OH)2 to CaCl2?

Yes, even if CaO reacts with water, later water is regenerated in the reaction of hydroxide with HCl. We are told amount of acid was stoichiometric.
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Rutherford

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Re: Problem of the week - 10/12/2012
« Reply #5 on: December 12, 2012, 12:45:33 AM »

I got that result when I used that the number of moles are 0.4 and 0.6mol (for the salts) n of HCl is therefore 2mol, and that the mass share is 3.5784% (4 digits). So:
73g:x=3.5784%:(100-3.5784)%, x=1967.02g of water.
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