Impurity Percentage?

Sodium chloride does not react with HCl.

Balanced equation for the reaction in titration:
2 HCl(aq) + Na₂CO₃(aq) → 2 NaCl(aq) + H₂O(l) + CO₂(g)
Mole ratio HCl : Na₂CO₃ = 2 : 1

No. of moles of HCl reacted in titration = (1 mol/L) × (27.3/1000 L) = 0.0273 mol
No. of moles of Na₂CO₃ = (0.0273 mol) × (1/2) = 0.01365 mol
No. of moles of Na₂CO₃ in 250 cm³ of the solution = (0.01365 mol) × (250/25) = 0.1365 mol
No. of moles of Na₂CO₃ in the impure sample = 0.1365 mol

Molar mass of Na₂CO₃ = (23.0×2 + 12.0 + 16.0×3) g/mol = 106.0 g/mol
Mass of Na₂CO₃ in the impure sample = (0.1365 mol) × (106.0 g/mol) = 14.469 g
Percentage by mass of Na₂CO₃ in the sample = (14.469/15.0) × 100% = 96.46%
Percentage by mass of NaCl impurity in the sample = (100 - 96.46)% = 3.54%


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OR:
Percentage by mass of NaCl impurity
= 100% - [(1 mol HCl / L) × (27.3/1000 L) × (1 mol Na₂CO₃ / 2 mol HCl) × (250/25) × (106.0 g / mol Na₂CO₃) / (15.0 g)] × 100%
= 3.54%

balanced equations
.. 1 Na2CO3 + 2 HCl ---> 2 NaCl + 1 H2O + 1 CO2
.. 1 NaCl + 1 HCl --> no reaction

note that ONLY the Na2CO3 reacts!

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using dimensional analysis to convert moles HCl to moles Na2CO3 to mass Na2CO3

.. 27.3cm³ HCl.. ... 1L HCl... .. . __ mol HCl.. 1 mol Na2CO3.. 106.0g Na2CO3
.. ---- ---- ---- --- x --- ---- --- --- x ---- ----- ----- x --- ---- --- ----- x --- ---- --- ---- ---- = __g Na2CO3
.. ... .. .. 1.. ... .. .. 1000mL HCl... .. 1 L HCl.. .... 2 mol HCl.. ... . 1 mol Na2CO3

and that is in that 25cm³ aliquot, so
.. mass Na2CO3 in the 250cm³ initial solution = mass Na2CO3 in the 15.0g sample = __ g

and finally
.. mass % NaCl in the sample = (15.0g - __g) / (15.0g) x 100% = ___ %

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fill in the blanks!