Question

Carbon dioxide gas, saturated with water vapor, can be produced by the addition of aqueous acid to calcium carbonate. \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) How many moles of \(\mathrm{CO}_{2}(g),\) collected at \(60 .^{\circ} \mathrm{C}\) and 774 torr total pressure, are produced by the complete reaction of \(10.0 \mathrm{g}\) of \(\mathrm{CaCO}_{3}\) with acid? What volume does this wet \(\mathrm{CO}_{2}\) occupy? What volume would the \(\mathrm{CO}_{2}\) occupy at 774 torr if a desiccant (a chemical drying agent) were added to remove the water? (The vapor pressure of water at \(60 .^{\circ} \mathrm{C}\) is \(149.4 \mathrm{mm} \mathrm{Hg} .)\)

Short answer

The number of moles of \(\mathrm{CO}_{2}\) produced is \(0.100\,\mathrm{mol}\). The volume of wet \(\mathrm{CO}_{2}(g)\) is approximately \(2.69\,\mathrm{L}\), and the volume of dry \(\mathrm{CO}_{2}(g)\) is approximately \(3.34\,\mathrm{L}\).

Step-by-step solution

1. Find the number of moles of \(\mathrm{CO}_{2}\) produced

Given the mass of \(\mathrm{CaCO}_{3}\) is \(10.0\, \mathrm{g}\), we first need to find the molar mass of \(\mathrm{CaCO}_{3}\): $$ \mathrm{M_{CaCO_{3}}} = \mathrm{M_{Ca}} + \mathrm{M_{C}} + 3\mathrm{M_{O}} = 40.08 + 12.01 + 3(16.00) = 100.09\, \mathrm{g\, mol^{-1}} $$ Now, we calculate the moles of \(\mathrm{CaCO}_{3}\): $$ \mathrm{n_{CaCO_{3}}} = \frac{\mathrm{mass}}{\mathrm{Molar\, mass}} = \frac{10.0\, \mathrm{g}}{100.09\, \mathrm{g\, mol^{-1}}} = 0.100\, \mathrm{mol} $$ Using the stoichiometry of the balanced chemical equation, one mole of \(\mathrm{CaCO}_{3}\) produces one mole of \(\mathrm{CO}_{2}\), we can conclude that the number of moles of \(\mathrm{CO}_{2}\) is also \(0.100\).

2. Calculate the volume of wet \(\mathrm{CO}_{2}\) gas

We will use the Ideal Gas Law (\(PV = nRT\)) to calculate the volume of \(\mathrm{CO}_{2}(g)\). First, let's convert the given pressure from torr to atm: $$ \mathrm{Conversion}: 1\,\mathrm{atm} = 760\,\mathrm{torr} \\ \mathrm{Pressure} = 774\,\mathrm{torr} \times \frac{1\,\mathrm{atm}}{760\,\mathrm{torr}} = 1.0184\,\mathrm{atm} $$ Now, convert the temperature from Degrees Celsius to Kelvin: $$ \mathrm{Temperature\, (in\, Kelvin)} = 60\, ^{\circ}\mathrm{C} + 273.15 = 333.15\, \mathrm{K} $$ Now we can use the Ideal Gas Law to find the volume of \(\mathrm{CO}_{2}(g)\): $$ PV = nRT \\ V = \frac{nRT}{P} = \frac{0.100\, \mathrm{mol} \times 0.0821\,\mathrm{L\, atm\, mol^{-1}\, K^{-1}} \times 333.15\, \mathrm{K}}{1.0184\, \mathrm{atm}} \approx 2.69\, \mathrm{L} $$ The volume of the wet \(\mathrm{CO}_{2}(g)\) is approximately \(2.69\,\mathrm{L}\).

3. Calculate the volume of dry \(\mathrm{CO}_{2}\) gas

To find the volume of dry \(\mathrm{CO}_{2}\) gas, we need to subtract the partial pressure of water vapor from the total pressure. The vapor pressure of water at \(60\,^{\circ}\mathrm{C}\) is \(149.4\,\mathrm{mm\, Hg}\). Convert this to atm: $$ \mathrm{Partial\, Pressure\, of\, water\, vapor} = 149.4\,\mathrm{mm\, Hg} \times \frac{1\, \mathrm{atm}}{760\,\mathrm{mm\, Hg}} \approx 0.1966\,\mathrm{atm} $$ Now, subtract the partial pressure of water vapor from the total pressure to get the partial pressure of \(\mathrm{CO}_{2}(g)\). $$ \mathrm{Partial\, Pressure\, of\, CO_{2}} = 1.0184\,\mathrm{atm} - 0.1966\,\mathrm{atm} \approx 0.8218\,\mathrm{atm} $$ Finally, use the Ideal Gas Law again to find the volume of dry \(\mathrm{CO}_{2}(g)\): $$ V = \frac{nRT}{P} = \frac{0.100\, \mathrm{mol} \times 0.0821\,\mathrm{L\, atm\, mol^{-1}\, K^{-1}} \times 333.15\, \mathrm{K}}{0.8218\, \mathrm{atm}} \approx 3.34\, \mathrm{L} $$ The volume of dry \(\mathrm{CO}_{2}(g)\) is approximately \(3.34\,\mathrm{L}\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is one of the fundamental principles of chemistry, particularly in the study of gases. It is expressed by the equation \(PV = nRT\), where:

• \(P\) is the pressure of the gas in atmospheres.
• \(V\) is the volume of the gas in liters.
• \(n\) is the number of moles of the gas.
• \(R\) is the universal gas constant, approximately 0.0821 L atm mol\(^{-1}\) K\(^{-1}\).
• \(T\) is the temperature in Kelvin.

This law helps in calculating any of these variables if the others are known, describing how gases behave under varying conditions. When you apply the Ideal Gas Law in chemical reactions, ensure that units are standardized, for example, converting temperature to Kelvin and pressure to atmospheres. By knowing the moles of a gas, its pressure, and temperature, you can easily determine its volume using these calculations.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. Each reaction can be represented by a balanced chemical equation, which indicates the proportion of reactants and products with their respective stoichiometric coefficients. In the problem given, calcium carbonate \((\mathrm{CaCO}_{3})\) reacts with acid to produce carbon dioxide \((\mathrm{CO}_{2})\), water, and a calcium ion.In stoichiometry, the coefficients show how many moles of each substance are involved in the reaction. Here, one mole of \(\mathrm{CaCO}_{3}\) yields one mole of \(\mathrm{CO}_{2}\). Understanding this relationship is crucial for quantifying the amount of products formed from known quantities of reactants. Each reaction is a puzzle where the balance between elements must be maintained.

Molar Mass

Molar mass is the mass of one mole of a substance, measured in grams per mole \(\mathrm{g/mol}\). It is essential for converting between mass and moles in chemical calculations. To find the molar mass of a compound like calcium carbonate \((\mathrm{CaCO}_{3})\), you sum the atomic masses of all its constituent elements:

• Calcium \(\mathrm{M_{Ca}} = 40.08 \mathrm{g/mol}\)
• Carbon \(\mathrm{M_{C}} = 12.01 \mathrm{g/mol}\)
• Oxygen \(\mathrm{3} \times \mathrm{M_{O}} = 48.00 \mathrm{g/mol}\)

Adding these gives the molar mass of \(\mathrm{CaCO}_{3}\), which is 100.09 \(\mathrm{g/mol}\). By knowing the molar mass, you can convert grams of a substance to moles, a critical skill in stoichiometry and the Ideal Gas Law.

Partial Pressure

Partial pressure refers to the pressure contributed by a single type of gas in a mixture of gases. Dalton’s Law of Partial Pressures states that the total pressure is equal to the sum of the partial pressures of all individual gases in the mixture. In the exercise, carbon dioxide \((\mathrm{CO}_{2})\) gas is collected in the presence of water vapor. The total pressure includes both the pressure from \(\mathrm{CO}_{2}\) and water vapor. The vapor pressure of water at a specific temperature, in this case, 60 °C with 149.4 mm Hg, is subtracted from the total pressure to find the partial pressure of \(\mathrm{CO}_{2}\). This adjustment is necessary for calculating the dry gas volume using the Ideal Gas Law, ensuring accurate measurements in experiments involving gaseous compounds.