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Topic: neutralization  (Read 4499 times)

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Offline kimi85

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neutralization
« on: May 07, 2008, 01:54:53 AM »
A sample consistng entirely of pure Li2CO3 and BaCO3 weighs 1.500g and
 requires 55.00mL of 0.3600 N HCl for neutralization. What is the number
 of grams of Li2CO3 in the sample? (mol wt.Li2CO3 = 73.89. Mol wt.
 BaCO3 = 197.35)

choices: A. 0.3592g B. 0.1543g C. 0.0837g D. 0.2716g E.None of the given answrs.

thank you very much

Offline Borek

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Re: neutralization
« Reply #1 on: May 07, 2008, 03:19:54 AM »
Please read forum rules.

Although with so many posts you were for sure already asked to do so. Show us some effort.
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Offline kimi85

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Re: neutralization
« Reply #2 on: May 07, 2008, 03:34:36 AM »
hi!

this is my solution:

 (N x mL) HCl = x  Li2CO3/mw Li2CO3  + (1.5 - x)/ mwBaCO3

Is that correct? Should I multiply the right side by 3? thank you

Offline Borek

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Re: neutralization
« Reply #3 on: May 07, 2008, 04:30:42 AM »
(N x mL) HCl = x  Li2CO3/mw Li2CO3  + (1.5 - x)/ mwBaCO3

Not bad, you seem to know the general idea. Why is there x on the left?

Quote
Should I multiply the right side by 3?

You need to account for stoichiometry - but why 3?
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Offline AWK

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Re: neutralization
« Reply #4 on: May 07, 2008, 04:57:39 AM »
Quote
(N x mL) HCl = x  Li2CO3/mw Li2CO3  + (1.5 - x)/ mwBaCO3

You should use volume in liters (dm3), and according to stoichiometry of reactions you have to use a factor of 1/2 before product of volume and concentration of HCl
AWK

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