Question

Using Wetlands to Treat Agricultural Waste Wetlands can play a significant role in removing fertilizer residues from rain runoff and groundwater. One way they do this is through denitrification, which converts nitrate ions to nitrogen gas: \(2 \mathrm{NO}_{3}^{-}(a q)+5 \mathrm{CO}(g)+2 \mathrm{H}^{+}(a q) \right-arrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)+5 \mathrm{CO}_{2}(g)\) Suppose \(200.0 \mathrm{g}\) of \(\mathrm{NO}_{3}^{-}\) flows into a swamp each day. a. What volume of \(\mathrm{N}_{2}\) would be produced at \(17^{\circ} \mathrm{C}\) and 1.00 atm if the denitrification process were complete? b. What volume of \(\mathrm{CO}_{2}\) would be produced? c. Suppose the gas mixture produced by the decomposition reaction is trapped in a container at \(17^{\circ} \mathrm{C} ;\) what is the density of the mixture, assuming \(P_{\text {total }}\) \(=1.00\) atm?

Short answer

b. What is the volume of CO2 produced? c. What is the density of the gas mixture trapped within the container? a. The volume of N2 produced is 38.555 L. b. The volume of CO2 produced is 193.47 L. c. The density of the gas mixture trapped within the container is 1.209 g/L.

Step-by-step solution

Moles of NO3-

We are given 200.0 g of NO3-. To find the moles, we need the molar mass of NO3-. The molar mass of NO3- is \((1 \times 14.01)+ (3 \times 16.00) = 62.01 \, \mathrm{g/mol}\). The moles of NO3- are: \(\text{moles of NO3-}= \frac{200.0 \, \mathrm{g}}{62.01 \, \mathrm{g/mol}} = 3.225 \, \mathrm{mol}\).

Moles of N2 and CO2

According to the balanced equation, 2 moles of NO3- produce 1 mole of N2 and 5 moles of CO2. Therefore, we can calculate the moles of N2 and CO2 produced: \(\text{moles of N2} = \frac{1}{2} \times 3.225 \, \mathrm{mol} = 1.6125 \, \mathrm{mol}\) \(\text{moles of CO2} = \frac{5}{2} \times 3.225 \, \mathrm{mol} = 8.0625 \, \mathrm{mol}\)

Volume of N2 and CO2

We're given the temperature (17°C) and pressure (1 atm). We need to convert the temperature to Kelvin (K): \(T = 17 + 273 = 290 \, \mathrm{K}\) Now, we can use the ideal gas law, \(PV = nRT\), to find the volume of N2 and CO2 at given conditions: For N2: \(V_\text{N2} = \frac{n_\text{N2}RT}{P} = \frac{1.6125 \, \mathrm{mol} \cdot 0.0821 \, \frac{\mathrm{L \cdot atm}}{\mathrm{mol \cdot K}} \cdot 290 \, \mathrm{K}}{1\, \mathrm{atm}} = 38.555 \, \mathrm{L}\) For CO2: \(V_\text{CO2} = \frac{n_\text{CO2}RT}{P} = \frac{8.0625 \, \mathrm{mol} \cdot 0.0821 \, \frac{\mathrm{L \cdot atm}}{\mathrm{mol \cdot K}} \cdot 290 \, \mathrm{K}}{1\, \mathrm{atm}} = 193.47 \, \mathrm{L}\)

Density of gas mixture

The total moles of gases produced are: \(n_\text{total} = n_\text{N2} + n_\text{CO2} = 1.6125 + 8.0625 = 9.675 \, \mathrm{mol}\) Total mass of gases produced: \(m_\text{total} = n_\text{total} \cdot M = 9.675 \, \mathrm{mol} \cdot 28.97 \, \frac{\mathrm{g}}{\mathrm{mol}} = 280.44 \, \mathrm{g}\) Total volume of gases produced: \(V_\text{total} = V_\text{N2} + V_\text{CO2} = 38.555 + 193.47 = 232.03 \, \mathrm{L}\) Finally, density of the gas mixture: \(\rho = \frac{m_\text{total}}{V_\text{total}} = \frac{280.44 \, \mathrm{g}}{232.03 \, \mathrm{L}} = 1.209 \, \frac{\mathrm{g}}{\mathrm{L}}\) a. The volume of N2 produced is 38.555 L. b. The volume of CO2 produced is 193.47 L. c. The density of the gas mixture trapped within the container is 1.209 g/L.

Key concepts

Wetland Chemistry

Wetlands are fascinating ecosystems that act as nature's water treatment systems. They clean up water by breaking down pollutants through biological, chemical, and physical processes. One standout process is denitrification, where microbes convert nitrate (NO3-) into nitrogen gas (N2), effectively removing excess nutrients. This is crucial in preventing water pollution from agricultural runoff, which often contains high levels of fertilizers. By breaking down these chemicals, wetlands help maintain water quality and ecosystem health.
Beyond chemical conversion, wetlands provide habitat for diverse life forms and help in flood control. They're crucial for environmental balance, making the study of wetland chemistry both intriguing and essential for ecological health.

Ideal Gas Law

To understand how gases behave under different conditions, scientists use the Ideal Gas Law. It's formulated as \(PV = nRT\), where:

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

This relation helps predict how a gas will expand, compress, or react under certain environmental conditions. For example, in the denitrification process within wetlands, the volumes of nitrogen and carbon dioxide produced are calculated using this law.
Understanding the Ideal Gas Law is vital for many scientific and engineering applications, making it a cornerstone of fundamental chemistry.

Nitrate Conversion

Nitrate conversion is a key part of the denitrification process. It involves the transformation of nitrate ions (NO3-) into nitrogen gas (N2), lessening the nutrient load in water bodies. This process requires specific conditions including the presence of carbon sources, microbes, and anoxic (oxygen-free) environments.
The balanced chemical equation for this reaction: \[2 \mathrm{NO}_{3}^{-} + 5 \mathrm{CO} + 2 \mathrm{H}^{+} \rightarrow \mathrm{N}_{2} + \mathrm{H}_{2} \mathrm{O} + 5 \mathrm{CO}_{2}\] Here, nitrates are reduced, and gases like CO2 and N2 are released, crucial for maintaining nutrient cycles in nature. Studying nitrate conversion is essential for managing water quality and preventing eutrophication, where excessive nutrients lead to harmful algal blooms.

Gas Density Calculation

Gas density is defined as the mass of a gas per unit volume. It's calculated using the formula: \(\rho = \frac{m}{V}\), where:

• \(\rho\) is the density
• \(m\) is the mass
• \(V\) is the volume

In context, understanding gas density helps in analyzing how gases resulting from reactions, such as those in wetlands, interact and behave. By finding the total moles and mass of gases like N2 and CO2 produced in a reaction, we can determine how dense the gas mixture will be at specific conditions.
Such calculations are critical in both environmental studies and industrial applications, where the behavior of gases plays an important role. Understanding gas densities ensures efficient processes and better environmental management.