Question

Carbon dioxide gas, in the dry state, may be produced by heating calcium carbonate. $$ \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What volume of \(\mathrm{CO}_{2},\) collected dry at \(55^{\circ} \mathrm{C}\) and a pressure of 774 torr, is produced by complete thermal decomposition of \(10.0 \mathrm{g}\) of \(\mathrm{CaCO}_{3} ?\)

Short answer

The volume of \(CO_2\) produced by the complete thermal decomposition of 10.0 g of \(CaCO_3\) at \(55^{\circ}C\) and a pressure of 774 torr is 2.67 L.

Step-by-step solution

Calculate the moles of CaCO3

Given the mass of \(CaCO_3\), we will calculate the number of moles using the molar mass. Mass of \(CaCO_3\) = 10.0 g Molar mass of \(CaCO_3\) = (40.08 + 12.01 + 3 * 16.00) g/mol = 100.09 g/mol Now, to find the moles: Moles of \(CaCO_3\) = \(\frac{\text{mass of } CaCO_3}{\text{molar mass of } CaCO_3}\)= \(\frac{10.0 \text{g}}{100.09 \text{g/mol}}\) = 0.1 mol

Determine the moles of CO2 produced

Using the balanced chemical equation, we can determine the moles of \(CO_2\) produced from the moles of \(CaCO_3\). \(CaCO_3(s) \rightarrow CaO(s) + CO_2(g)\) From the balanced chemical equation, we see that 1 mole of \(CaCO_3\) produces 1 mole of \(CO_2\). Since we have 0.1 mole of \(CaCO_3\), we will have: Moles of \(CO_2\) = 0.1 mol

Calculate the volume of CO2

Now that we have the number of moles of \(CO_2\), we can use the ideal gas law equation, \(PV = nRT\), to calculate the volume. The given pressure and temperature are 774 torr and \(55 ^{\circ} C\), respectively. Before using the gas law equation, we need to convert pressure to atm and temperature to Kelvin: Pressure = \(\frac{774 \text{ torr}}{760 \text{ torr/atm}}\) = 1.018 atm Temperature = 55 + 273.15 = 328.15 K Now, let's use the ideal gas law equation, \(PV = nRT\), to solve for the volume (V). We will use the Universal Gas Constant \(R = 0.0821 \frac{\text{L atm}}{\text{mol K}}\) 1.018 atm x V = 0.1 mol x 0.0821 \(\frac{\text{L atm}}{\text{mol K}}\) x 328.15 K Solving for V: V = \(\frac{0.1 \text{ mol} \times 0.0821 \frac{\text{L atm}}{\text{mol K}} \times 328.15 \text{K}}{1.018 \text{ atm}}\) = 2.67 L So, the volume of \(CO_2\) produced by the complete thermal decomposition of 10.0 g of \(CaCO_3\) is 2.67 L.

Key concepts

Thermal Decomposition

Thermal decomposition is a chemical reaction where a single substance breaks down into two or more simpler substances when heated. In the case of calcium carbonate (CaCO_3), applying heat causes it to decompose into calcium oxide (CaO) and carbon dioxide (CO_2) gas. This reaction can be represented as follows:\[ \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \]This process is essential not only in laboratory settings but also in various industrial applications, such as cements production. When performing this reaction, it's crucial to ensure that our collected gas, in this case, carbon dioxide, is measured under controlled conditions. The temperature and pressure must be correctly accounted for, as they impact the gas volume significantly. Understanding thermal decomposition reactions is fundamental for analyzing how different compounds react upon heating and predicting the products formed.

Stoichiometry

Stoichiometry involves calculating the relationships between reactants and products in a chemical reaction. This field is crucial when dealing with balanced chemical equations, as they show the exact proportion of reactants and products needed for the reaction. Consider our balanced equation for thermal decomposition:\[ \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \]According to the equation, one mole of CaCO_3 produces one mole of CO_2. This direct one-to-one relationship enables us to predict the amount of product generated from a known quantity of reactant. Stoichiometry is essential for both calculating the required amount of reactants and minimizing waste in chemical reactions. By fully understanding stoichiometry, students can accurately conduct experiments and scale chemical processes for industrial applications.

Moles Calculation

Calculating moles is a fundamental skill in chemistry that helps in converting measurements into understandable quantities of substances. Moles express an amount of substance and are linked to the atomic/molecular scale. We can calculate moles with the formula:Moles = \( \frac{\text{Mass}}{\text{Molar Mass}} \)In the exercise, given 10.0 grams of CaCO_3 and its molar mass of 100.09 g/mol, the moles are calculated as follows:\[ 0.1 \text{ mol} = \frac{10.0 \text{ g}}{100.09 \text{ g/mol}} \]Understanding this conversion is crucial, as it forms the basis for further calculations in the chemical reaction, such as predicting amounts of products or the required volume for gaseous products. Mastery of mole calculations ensures accurate quantitative analysis in experiments and real-world applications.

Gas Volume Calculation

Gas volume calculations involve understanding the behavior of gases under various conditions. The ideal gas law is a powerful tool used here, which relates the pressure, volume, temperature, and moles of a gas. The equation is:\[ PV = nRT \]Where:

• \(P\) is the pressure in atmospheres.
• \(V\) is the volume in liters.
• \(n\) is the number of moles of gas.
• \(R\) is the universal gas constant \(0.0821 \frac{\text{L atm}}{\text{mol K}}\).
• \(T\) is the temperature in Kelvin.

In this particular example, the given conditions are 774 torr pressure (converted to 1.018 atm) and a temperature of 55°C (converted to 328.15 K). Using the calculated moles of CO_2, the volume is derived as:\[ V = \frac{0.1 \text{ mol} \times 0.0821 \frac{\text{L atm}}{\text{mol K}} \times 328.15 \text{ K}}{1.018 \text{ atm}} = 2.67 \text{ L} \]Gas volume calculations allow chemists to predict how a gas will behave in different scenarios, which is crucial for both experimental and industrial gas handling and applications.