Question

When 1.25 g of sample of chalk is strongly heated, 0.44 g of $CO_2$ gas is produced. Determine Percentage of pure $CaCO_3$ in chalk?

Answer 1

80%

Answer 2

The thermal decomposition of calcium carbonate ($CaCO_3$) is given by the reaction: $CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)$

The molar masses are: $M(CaCO_3) = 100$ g/mol and $M(CO_2) = 44$ g/mol. From stoichiometry, 100 g of $CaCO_3$ produces 44 g of $CO_2$.

To produce 0.44 g of $CO_2$, the mass of pure $CaCO_3$ required is: $m_{CaCO_3} = 0.44 \text{ g } CO_2 \times \frac{100 \text{ g } CaCO_3}{44 \text{ g } CO_2} = 1 \text{ g } CaCO_3$

The percentage of pure $CaCO_3$ in the chalk sample is calculated as: $\% \text{ pure } CaCO_3 = \frac{\text{Mass of pure } CaCO_3}{\text{Mass of sample}} \times 100$ $\% \text{ pure } CaCO_3 = \frac{1 \text{ g}}{1.25 \text{ g}} \times 100 = 80\%$