Posted by siari ann on Thursday, February 14, 2013 at 3:19am.
I assume you titrated ALL of the Na2CO3; i.e., Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
mols HCl = M x L = 3.00 x 0.0658 = approximately 0.2 (you can do it more accurately).
mols Na2CO3 = 1/2 that (from the coefficients) = approximately 0.1
g Na2CO3 = mols x molar mass = 0.1 x 106 = approximately 10 g.
mass of the 50 mL soln = 50 x 1.25 = 62.5 grams.
% Na2CO3 = (10/62.5)*10 = ?%
..
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