Understanding Thermal Decomposition
When calcium carbonate ($ \text{CaCO}_3$) is heated, it undergoes a thermal decomposition reaction to form calcium oxide (quicklime) and carbon dioxide gas. The balanced chemical equation for this process is:
$\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$
This reaction shows a 1:1 molar ratio between the reactant $\text{CaCO}_3$ and the product $\text{CO}_2$.
Step-by-Step Problem Solving
To find the volume of $\text{CO}_2$ produced, we need to follow these logical steps:
1. Determine the pure amount of $\text{CaCO}_3$
We are given 20g of a sample that is only 20% pure. We must find the actual mass of $\text{CaCO}_3$ available for the reaction:
$\text{Mass of pure } \text{CaCO}_3 = 20\text{ g} \times \frac{20}{100} = 4\text{ g}$
2. Calculate the Molar Mass of $\text{CaCO}_3$
- Ca: 40 g/mol
- C: 12 g/mol
- O: 3 \times 16 = 48 g/mol
- Total: $40 + 12 + 48 = 100\text{ g/mol}$
3. Convert Mass to Moles
Using the formula $n = \frac{m}{M}$:
$n(\text{CaCO}_3) = \frac{4\text{ g}}{100\text{ g/mol}} = 0.04\text{ moles}$
4. Relate Moles to $\text{CO}_2$
According to the stoichiometry of the reaction, 1 mole of $\text{CaCO}_3$ produces 1 mole of $\text{CO}_2$. Therefore, 0.04 moles of $\text{CaCO}_3$ will produce 0.04 moles of $\text{CO}_2$.
5. Calculate Volume of $\text{CO}_2$ at STP
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies 22.4 Liters.
$\text{Volume of } \text{CO}_2 = 0.04\text{ mol} \times 22.4\text{ L/mol} = 0.896\text{ Liters}$
Conclusion
When 20g of 20% pure $\text{CaCO}_3$ is completely heated, it produces 0.896 Liters (or 896 mL) of $\text{CO}_2$ gas at STP.