Question

10 g sample of CaCl2 and NaCl is treated to precipitate all the Calcium as CaCO3. This CaCO3 is heated to convert all the Ca into CaO and final mass of CaO is 1.62 g. The % by mass of CaCl2 in the original mixture is?

Answer 1

Question#47294 – Chemistry – Inorganic Chemistry

Question:

10 g sample of $\mathrm{CaCl}_2$ and NaCl is treated to precipitate all the Calcium as $\mathrm{CaCO}_3$. This $\mathrm{CaCO}_3$ is heated to convert all the Ca into CaO and final mass of CaO is 1.62 g. The % by mass of $\mathrm{CaCl}_2$ in the original mixture is?

Answer:

The precipitation of $\mathrm{CaCO}_3$:

$$\mathrm{CaCl}_2 + \mathrm{CO}_3^{2-} \rightarrow \mathrm{CaCO}_3 + 2\mathrm{Cl}^-$$

Calcium chloride will react, while sodium chloride will remain untached.

The heating of $\mathrm{CaCO}_3$:

$$\mathrm{CaCO}_3 \rightarrow \mathrm{CaO} + \mathrm{CO}_2$$

If the mass of CaO produced is 1.62 g, than the corresponding mass of $\mathrm{CaCO}_3$:

$$m(\mathrm{CaCO}_3) = \frac{m(\mathrm{CaO}) \times M(\mathrm{CaCO}_3)}{M(\mathrm{CaO})} = \frac{1.62\,\mathrm{g} \times 100\,\mathrm{g/mol}}{56\,\mathrm{g/mol}} = 2.89\,\mathrm{g}$$

The corresponding mass of $\mathrm{CaCl}_2$:

$$m(\mathrm{CaCl}_2) = \frac{m(\mathrm{CaCO}_3) \times M(\mathrm{CaCl}_2)}{M(\mathrm{CaCO}_3)} = \frac{2.89\,\mathrm{g} \times 111\,\mathrm{g/mol}}{100\,\mathrm{g/mol}} = 3.21\,\mathrm{g}$$

The percentage of $\mathrm{CaCl}_2$:

$$\omega(\mathrm{CaCl}_2) = \frac{m(\mathrm{CaCl}_2)}{m(\mathrm{CaCl}_2) + m(\mathrm{NaCl})} \times 100\% = \frac{3.21}{10} \times 100\% = 32.1\%$$