Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Using Solutions in Chemical Reactions - Page 111: 83

The purity of sodium carbonate is $61.89\%$.

Considering stoichiometric coefficients, we can conclude that $n(Na_2CO_3) : n(HCl)=1:2$, hence $n(Na_2CO_3)=\frac{1}{2}n(HCl)=\frac{1}{2}\times c(HCl)\times V(HCl)=\frac{1}{2}\times 0.1755\frac{mol}{dm^3}\times 0.01555dm^3=1.36\times 10^{-3}mol$ The mass of pure $Na_2CO_3$ can now be easily obtained: $m(Na_2CO_3)=M(Na_2CO_3)\times n(Na_2CO_3)=106\frac{g}{mol}\times 1.36\times 10^{-3}mol=0.1446g$ Therefore, the purity of sodium carbonate is $\frac{0.1446g}{0.2337g}=61.89\%$