[kimi85]

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

[Borek]

Please read forum rules.

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

[kimi85]

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

[Borek]

(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?

Should I multiply the right side by 3?

You need to account for stoichiometry - but why 3?

[AWK]

(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