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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: kimi85 on May 07, 2008, 01:54:53 AM

Title: neutralization
Post by: kimi85 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
Title: Re: neutralization
Post by: Borek on May 07, 2008, 03:19:54 AM
Please read forum rules (http://www.chemicalforums.com/index.php?page=forumrules).

Although with so many posts you were for sure already asked to do so. Show us some effort.
Title: Re: neutralization
Post by: kimi85 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
Title: Re: neutralization
Post by: Borek on May 07, 2008, 04:30:42 AM
Quote from: kimi85 on May 07, 2008, 03:34:36 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?
Title: Re: neutralization
Post by: AWK 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