Sodium carbonate of 92% purity is used in the reaction Na₂CO₃ + CaCl₂ → CaCO₃ + 2NaCl. The number of grams of Na₂CO₃ required to yield one gram of CaCO₃

$$\begin{gathered} {\mathrm{Na}}_2{\mathrm{CO}}_3 + {\mathrm{CaCl}}_2 \to {\mathrm{CaCO}}_3 + 2\mathrm{NaCl} \\ 106\mathrm{g} 100\mathrm{g} \\ ? 1\mathrm{g} \\ \mathrm{Mass} \mathrm{of} \mathrm{pure} {\mathrm{Na}}_2{\mathrm{CO}}_3=\frac{106}{100}=1.06\mathrm{g} \\ \mathrm{Mass} \mathrm{of} 92\% {\mathrm{Na}}_2{\mathrm{CO}}_3=\frac{1.06}{92}\times 100=1.152\mathrm{g} \end{gathered}$$