Question

a sample of CaCO3 of mass 2.8g was reacted with 75 cm3 of 1.0 mol/dm3 hydrochloride acid. The resulting solution was completely transferred into 250 cm3 of volumetric flask and the mixture made up to the mark with distilled water. 25 cm3 of this solution needed 20.05 cm3 of 1.0 mol/dm3 of NaOH to neutralize the excess acid. Calculate the percentage of CaCO3 present in the sample.

Answer 1

Answer on Question #85043 – Chemistry – Other

Task:

A sample of $\mathrm{CaCO_3}$ of mass $2.8\mathrm{g}$ was reacted with $75~\mathrm{cm}^3$ of $1.0\mathrm{mol} / \mathrm{dm}^3$ hydrochloride acid. The resulting solution was completely transferred into $250~\mathrm{cm}^3$ of volumetric flask and the mixture made up to the mark with distilled water. $25~\mathrm{cm}^3$ of this solution needed $20.05~\mathrm{cm}^3$ of $1.0\mathrm{mol} / \mathrm{dm}^3$ of NaOH to neutralize the excess acid. Calculate the percentage of $\mathrm{CaCO_3}$ present in the sample.

Solution:

Reaction of neutralization of excess hydrochloric acid:

$$HCl + NaOH = NaCl + H_2O$$

According to the chemical reaction equation:

$$\begin{array}{l} n(HCl) = n(NaOH) \\ n(HCl) = C(NaOH) * V(NaOH) \end{array}$$

Then,

$$\begin{array}{l} n_{\text{excess}}(HCl) = 1.0 \, \mathrm{mol} / \mathrm{dm}^3 * 0.02005 \, \mathrm{dm}^3 = 0.02005 \, \mathrm{mol} \\ n_{\text{excess}}(HCl) = 0.02005 \, \mathrm{mol} \end{array}$$

Correcting for the aliquoting/dilution factor, you have:

$$\begin{array}{l} n_{\text{excess}}^0(HCl) = \frac{V_f}{V_a} * n_{\text{excess}}(HCl) = \frac{250 \, \mathrm{cm}^3}{25 \, \mathrm{cm}^3} * 0.02005 \, \mathrm{mol} = 0.2005 \, \mathrm{mol} \\ n_{\text{excess}}^0(HCl) = 0.2005 \, \mathrm{mol} \end{array}$$ $$\begin{array}{l} n_{\text{total}}(HCl) = C(HCl) * V_{\text{total}}(HCl) = 1.0 \, \mathrm{mol} / \mathrm{dm}^3 * 0.075 \, \mathrm{dm}^3 = 0.075 \, \mathrm{mol} \\ n_{\text{total}}(HCl) < n_{\text{excess}}^0(HCl) = 0.075 \, \mathrm{mol} < 0.2005 \, \mathrm{mol} \end{array}$$

There is a error with the way the question is presented because according to the above calculations, there is more HCl left over (in excess) than there was originally added.

$$n(HCl) = n_{\text{total}}(HCl) - n_{\text{excess}}^{0}(HCl)$$

Reaction of $\mathrm{CaCO_3}$ with hydrochloric acid:

$$\mathrm{CaCO_3} + 2HCl = \mathrm{CaCl_2} + H_2O + CO_2$$

According to the chemical reaction equation:

$$n(\mathrm{CaCO_3}) = \frac{n(HCl)}{2} = \frac{n_{\text{total}}(HCl) - n_{\text{excess}}^{0}(HCl)}{2};$$ $$m(\mathrm{CaCO_3}) = n(\mathrm{CaCO_3}) \cdot M(\mathrm{CaCO_3}) = \left( \frac{n_{\text{total}}(HCl) - n_{\text{excess}}^{0}(HCl)}{2} \right) \cdot M(\mathrm{CaCO_3})$$

Then,

$$m''(\mathrm{CaCO_3}) = 2.8g$$ $$\% \text{yield} = \frac{m(\mathrm{CaCO_3})}{m''(\mathrm{CaCO_3})} \cdot 100\% = \frac{m(\mathrm{CaCO_3})}{2.8g} \cdot 100\%$$

**Answer:** Error in the task. Problem is that more HCl left over (0.2005 moles) than was originally present (0.075 moles).

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