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?

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 find the percentage by mass of NaCl in the original mixture of NaCl and Na2CO3, we need to determine the amount of NaCl present in the mixture and calculate it as a percentage of the total mass.

Let's break down the problem step by step:

1. Determine the moles of HCl used in the reaction:
Moles of HCl = Molarity × Volume (in liters)
Moles of HCl = 0.243 M × 0.055 L
Moles of HCl = 0.013365 moles

2. Balance the chemical equation for the reaction between HCl and Na2CO3:
2 HCl + Na2CO3 → 2 NaCl + H2O + CO2

From the balanced equation, we can see that two moles of HCl are required to react with one mole of Na2CO3 to produce two moles of NaCl. Therefore, the moles of NaCl formed in the reaction are also 0.013365 moles.

3. Determine the moles of NaCl in the original mixture:
Since the moles of NaCl formed in the reaction are equal to the moles of NaCl in the original mixture, the mole ratio is 1:1.

Moles of NaCl in the original mixture = 0.013365 moles

4. Calculate the mass of NaCl in the original mixture:
Mass of NaCl = Moles of NaCl × Molar Mass of NaCl

The Molar Mass of NaCl is 58.44 g/mol.

Mass of NaCl = 0.013365 moles × 58.44 g/mol
Mass of NaCl = 0.780 g

5. Calculate the percentage by mass of NaCl:
Percentage by mass = (Mass of NaCl / Total mass of the mixture) × 100%

Total mass of the mixture is given as 1.300 g.

Percentage by mass of NaCl = (0.780 g / 1.300 g) × 100%
Percentage by mass of NaCl = 60.0%

Therefore, the percentage by mass of NaCl in the original mixture of NaCl and Na2CO3 is 60.0%.