I think the easy way to solve this problem is to recognize that the HCl titrates the Na2CO3 and leaves the NaCl as is.
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
Total HCl added = M x L = approx 13.365 millimols but you need to confirm all of these numbers since I'm estimated some of them. Then some of the HCl (the unused amount) was 7.4 mL x 0.1M = 0.74 millimols excess HCl. So the amount of HCl actually used in the reaction was 13.365-0.74 = approx 12.625 mmols or 0.012625 mols.
Convert mols HCl to mols Na2CO3 using the coefficients in the balanced equation. That will be 1/2 mols HCl.
Convert that to grams Na2CO3. That's mols Na2CO3 x molar mass Na2CO3.
Then g NaCl = total - g Na2CO3
%NaCl = (g NaCl/1.300)*100 = ?
To a mixture of NaCl and Na2CO3 with a mass of 1.300 g was added 55.00 mL of 0.243 M HCl (an excess of HCl). The mixture was warmed to expel all of the CO2 and then the unreacted HCl was titrated with 0.100 M NaOH. The titration required 7.40 mL of the NaOH solution. What was the percentage by mass of NaCl in the original mixture of NaCl and Na2CO3?
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