The titration reaction of a solution containing 1.00 g mixture of Na2CO3 and K2CO3 with HCl shows that there are 8.50x10^-3 moles of C03-2. Find the weight percent of Na2CO3 and K2CO3 in the mixture. Please Help!!

Answer 1

Two equation solved simultaneously. You follow.
Let x = mass Na2CO3
and y = mass K2CO3
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equation 1 is x + y = 1.00 g

That's the easy one. What you do for equation 2 is to say mols CO3^2- from Na2CO3 + mols CO3^2- from K2CO3 = 8.50E-3 mols CO3^2-. Now we put that simple statement into chemical terms using x and y. To save typing space let's let MM = molar mass.
The mols CO3^2- from Na2CO3 =
(x/MM Na2CO3) and mols CO3^2- from K2CO3 = (y/MM K2CO3) so equation 2 is
(x/MM Na2CO3) + (y/MM K2CO3) = 8.50E-3

Solve equations 1 and 2 simultaneously for x and y, then mass percent is
%Na2CO3 = (x/mass sample)*100 = ?
%K2CO3 = (y/mass sample)*100 = ?