Question

When calcium carbonate is heated strongly, carbon dioxide gas is evolved. $$ \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ If \(4.74 \mathrm{g}\) of calcium carbonate is heated, what volume of \(\mathrm{CO}_{2}(g)\) would be produced when collected at \(26^{\circ} \mathrm{C}\) and 0.997 atm?

Short answer

The volume of carbon dioxide gas produced when 4.74 g of calcium carbonate is heated at 26°C and 0.997 atm is 1.16 L.

Step-by-step solution

Calculate moles of calcium carbonate

First, we need to calculate the moles of calcium carbonate. We'll use the molar mass of calcium carbonate (\(CaCO_3\)) and the given mass of \(CaCO_3\). Molar mass of \(CaCO_3= 40.08 (Ca) + 12.01 (C) + (3\times16.00) (O)= 100.09\: g/mol\) Moles of \(CaCO_3 = \frac{mass}{molar\:mass} = \frac{4.74 g}{100.09\ g/mol} = 0.04735\:mol\)

Determine the mole ratio of calcium carbonate to carbon dioxide

The balanced reaction equation for the decomposition of calcium carbonate is: $$ \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ The mole ratio between \(CaCO_3\) and \(CO_2\) is 1:1.

Calculate moles of carbon dioxide formed

Since the mole ratio between \(CaCO_3\) and \(CO_2\) is 1:1, the moles of carbon dioxide formed are the same as the moles of calcium carbonate. Moles of \(CO_2 = 0.04735\:mol\)

Calculate the volume of carbon dioxide gas

Now, we'll use the ideal gas law to calculate the volume of carbon dioxide gas produced. The ideal gas law is given by: $$ PV = nRT $$ Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We're given the pressure (0.997 atm) and temperature (26°C = 299 K), and we calculated the moles of \(CO_2\) in the previous step. The ideal gas constant, \(R = 0.0821\frac{L\times atm}{mol\times K}\). Rearrange the ideal gas law to solve for the volume, V: $$ V = \frac{nRT}{P} $$ Substitute the given values: $$ V = \frac{(0.04735\:mol)(0.0821\frac{L\times atm}{mol\times K})(299\:K)}{ (0.997\:atm)} $$ Finally, calculate the volume of carbon dioxide gas: $$ V = 1.16\:L $$ The volume of carbon dioxide gas produced when 4.74 g of calcium carbonate is heated at 26°C and 0.997 atm is 1.16 L.

Key concepts

Stoichiometry

Understanding stoichiometry is akin to being a chef following a recipe; it tells you how much of each ingredient you need for the desired product. In chemistry, ingredients are reactants, and our product is the outcome of a chemical reaction.

Stoichiometry is based on the balanced chemical equation, which adheres to the Law of Conservation of Mass. For the reaction where calcium carbonate (\texttt{CaCO3}) decomposes into calcium oxide (\texttt{CaO}) and carbon dioxide (\texttt{CO2}), the balanced equation is:
$$\texttt{CaCO3}(s) \rightarrow \texttt{CaO}(s)+\texttt{CO2}(g)$$
This indicates a 1:1:1 mole ratio. To determine how much \texttt{CO2} is produced from a given amount of \texttt{CaCO3}, we apply this mole ratio. In other words, 1 mole of \texttt{CaCO3} will yield 1 mole of \texttt{CO2} when decomposed. This is the fundamental concept of stoichiometry: it allows us to predict the quantity of a product formed from a known amount of reactant.

Ideal Gas Law

For gases, the ideal gas law serves as a universal recipe linking pressure, volume, temperature, and number of moles. The mathematical representation is:$$PV = nRT$$
where:

• \texttt{P} is the pressure in atmospheres,
• \texttt{V} is the volume in liters,
• \texttt{n} is the number of moles of the gas,
• \texttt{R} is the ideal gas constant (\texttt{0.0821 L·atm/mol·K}),
• \texttt{T} is the temperature in Kelvin.

To convert from Celsius to Kelvin, simply add 273 to the Celsius temperature. This law assumes gases behave ideally, meaning they have no intermolecular forces and occupy no volume themselves, which is an excellent approximation for many gases at standard temperatures and pressures.

For the given problem, we manipulate this law to solve for the volume of carbon dioxide gas. By inserting the number of moles (\texttt{n}), the constant (\texttt{R}), the temperature in Kelvin (\texttt{T}), and the pressure (\texttt{P}) into the equation, we calculate the volume (\texttt{V}) of \texttt{CO2} produced.

Molar Mass

Molar mass is the weight of one mole of a substance, typically expressed in grams per mole. It's essential to grasp this concept, as it acts as a conversion factor between grams and moles, vital in stoichiometric calculations.

To calculate the molar mass, you add the atomic masses of each element in a compound. For calcium carbonate (\texttt{CaCO3}), we sum the atomic masses of calcium (\texttt{Ca}), carbon (\texttt{C}), and three oxygens (\texttt{O}) to get about \texttt{100.09 g/mol}:
$$\texttt{Molar mass of CaCO3 = 40.08 (Ca) + 12.01 (C) + (3 \times 16.00) (O)}$$
Thus, knowing the mass of \texttt{CaCO3} allows us to find the moles by dividing the mass by the molar mass. With the number of moles, we can travel through stoichiometry and the ideal gas law to end at the volume of gas produced, as exemplified in this exercise.