Question

10g of sample of mixture of cacl2 and nacl is treated to precipitate all calcium as caco3 this caco3 is heated to convert all ca to cao and final mass of cao is 1.62g. The percent by mass of cacl2 in original mixture is

Answer 1

Answer on Question #48966 - Chemistry – Inorganic Chemistry

Question

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

Answer:

The scheme of this converting is:

$$\mathrm{CaCl}_2 \rightarrow \mathrm{CaCO}_3 \rightarrow \mathrm{CaO}$$

We see that 1 mole of $\mathrm{CaCl}_2$ forms 1 mole of $\mathrm{CaCO}_3$, then 1 mole of $\mathrm{CaCO}_3$ forms 1 mole of CaO.

Number of moles of CaO is:

$$n(\mathrm{CaO}) = \frac{m(\mathrm{CaO})}{M(\mathrm{CaO})} = \frac{1.62}{40.1} = 0.029 \text{ mol}$$

Therefore, number of moles of $\mathrm{CaCl}_2$ in original mixture was 0.029 moles too. Then the mass of $\mathrm{CaCl}_2$ in original mixture is:

$$m(\mathrm{CaCl}_2) = n(\mathrm{CaCl}_2)M(\mathrm{CaCl}_2) = 0.029 \cdot 111.1 = 3.22 \text{ g}$$

The percent by mass of $\mathrm{CaCl}_2$ in original mixture is:

$$\omega(\mathrm{CaCl}_2) = \frac{m(\mathrm{CaCl}_2)}{m(\text{mixture})} \cdot 100\% = \frac{3.22}{10} \cdot 100\% = 32.2\%$$

Answer: 32.2 %

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