Question

Chalk is impure calcium carbonate. The amount of calcium carbonate present can be determined by hydrochloric acid to a sample of chalk and measuring the volume of carbon dioxide produced CaCo3(aq) + 2HCl -> CaCl2(aq) + Co2(g) + H2O(g) Excess hydrochloric acid was added to 0.5g chalk and 100cm3 of carbon dioxide gas was given produced at r.t.p calculate the percentage purity of calcium carbonate in sample of chalk

Answer 1

Approximately
`83\%`.

Step-by-step explanation:

How many moles of
`\mathrm{CO_2}` gas are released?

The volume of each mole of of an ideal gas is approximately
`\rm 24\;L` under room temperature and pressure (r.t.p,
`\rm 20\;^(\circ)C`,
`\rm 1\; \right. atm`.) That's the same as
`\rm 24,000\;cm^(2)`.

Assume that
`\mathrm{CO_2}` acts like an ideal gas.

`\displaystyle n(\mathrm{CO_2}) = \rm (100\; cm^(3))/(24,000\; cm^(3)) \approx 0.00416667\; mol`.

`\rm HCl` is in excess. How many moles of
`\mathrm{CaCO_3}` formula units will produce that
`\rm 0.00416667\; mol` of
`\mathrm{CO_2}`?

Consider the ratio between the coefficient of
`\mathrm{CaCO_3}` and that of
`\mathrm{CO_2}`.

`\displaystyle \frac{n(\mathrm{CaCO_3})}{n(\mathrm{CO_2})}=1`.

In other words,

`\displaystyle n(\mathrm{CaCO_3})= n(\mathrm{CO_2})\cdot \frac{n(\mathrm{CaCO_3})}{n(\mathrm{CO_2})} = \rm 0.00416667\; mol`.

What's the mass of that many
`\mathrm{CaCO_3}`?

Relative atomic mass data from a modern periodic table:

• Ca: 40.078;
• C: 12.011;
• O: 15.999.

Formula mass of
`\mathrm{CaCO_3}`:

`M(\mathrm{CaCO_3}) = 40.078+12.011 + 3* 15.999 =\rm 100.086\; g\cdot mol^(-1)`.

Mass of that
`\rm 0.00416667\; mol` of
`\mathrm{CaCO_3}`:

`m = n \cdot M = \rm 0.00416667 * 100.086 = 0.417025\; g`.

Percentage mass of
`\mathrm{CaCO_3}` in this sample of chalk:

`\displaystyle \frac{\text{Mass of }\mathrm{CaCO_3}}{\text{Mass of Chalk}} * 100\%= \rm (0.417025\; g)/(0.5\; g) * 100\%\approx 83\%`.