Question

Ребята, помогите с задачей пожалуйста))))))) Даю 70 баллов за решение) Природный газ одного из месторождений содержит метан, объёмная доля его 92%, этан 3%, пропан 1,6%, бутан 0,4%, азот(N) 2%, оксид углерода(|\/), пары воды и другие валентные газы 1%. Какой объём воздуха потребуется для сжигания газа объёмом 5м^3. Объёмная доля (О)кислорода в воздухе составляет 21%

Answer 1

Решение вприложенном файле

Answer 2

Calculation of Air Volume Required for Combustion

To calculate the volume of air required for the combustion of the given gas, we need to determine the volume of oxygen (O2) needed for complete combustion.

The composition of the natural gas is as follows: - Methane (CH4): 92% - Ethane (C2H6): 3% - Propane (C3H8): 1.6% - Butane (C4H10): 0.4% - Nitrogen (N2): 2% - Carbon Dioxide (CO2): Unknown - Water Vapor and Other Valence Gases: 1%

The volume of air required for combustion can be calculated by considering the stoichiometry of the combustion reaction. The balanced equation for the combustion of methane (CH4) is as follows:

CH4 + 2O2 → CO2 + 2H2O

From the balanced equation, we can see that one mole of methane requires two moles of oxygen for complete combustion.

To calculate the volume of oxygen required, we need to convert the volume of natural gas (5 m^3) to moles. Since we know the volume percentages of each component in the natural gas, we can assume that the volume of each component is proportional to its mole fraction.

Let's calculate the moles of methane (CH4) in 5 m^3 of natural gas: - Volume of methane (CH4) = 92% of 5 m^3 - Moles of methane (CH4) = (Volume of methane / molar volume of methane)

Similarly, we can calculate the moles of other components in the natural gas.

Next, we need to calculate the moles of oxygen required for the combustion of the calculated moles of methane.

Finally, we can convert the moles of oxygen to volume using the molar volume of oxygen.

Calculation Steps:

1. Calculate the moles of methane (CH4) in 5 m^3 of natural gas: - Volume of methane (CH4) = 92% of 5 m^3 - Moles of methane (CH4) = (Volume of methane / molar volume of methane)

2. Calculate the moles of other components in the natural gas: - Moles of ethane (C2H6) = (Volume of ethane / molar volume of ethane) - Moles of propane (C3H8) = (Volume of propane / molar volume of propane) - Moles of butane (C4H10) = (Volume of butane / molar volume of butane) - Moles of nitrogen (N2) = (Volume of nitrogen / molar volume of nitrogen) - Moles of carbon dioxide (CO2) = (Volume of carbon dioxide / molar volume of carbon dioxide) - Moles of water vapor and other valence gases = (Volume of water vapor and other valence gases / molar volume of water vapor and other valence gases)

3. Calculate the total moles of oxygen required for combustion: - Moles of oxygen = (2 * Moles of methane) + (3 * Moles of ethane) + (4 * Moles of propane) + (5 * Moles of butane)

4. Convert the moles of oxygen to volume: - Volume of oxygen = (Moles of oxygen * molar volume of oxygen)

5. The calculated volume of oxygen is the same as the volume of air required for combustion, as air contains approximately 21% oxygen.

Let's perform the calculations:

Step 1: Calculate the moles of methane (CH4) in 5 m^3 of natural gas: - Volume of methane (CH4) = 92% of 5 m^3 - Moles of methane (CH4) = (Volume of methane / molar volume of methane)

Step 2: Calculate the moles of other components in the natural gas: - Moles of ethane (C2H6) = (Volume of ethane / molar volume of ethane) - Moles of propane (C3H8) = (Volume of propane / molar volume of propane) - Moles of butane (C4H10) = (Volume of butane / molar volume of butane) - Moles of nitrogen (N2) = (Volume of nitrogen / molar volume of nitrogen) - Moles of carbon dioxide (CO2) = (Volume of carbon dioxide / molar volume of carbon dioxide) - Moles of water vapor and other valence gases = (Volume of water vapor and other valence gases / molar volume of water vapor and other valence gases)

Step 3: Calculate the total moles of oxygen required for combustion: - Moles of oxygen = (2 * Moles of methane) + (3 * Moles of ethane) + (4 * Moles of propane) + (5 * Moles of butane)

Step 4: Convert the moles of oxygen to volume: - Volume of oxygen = (Moles of oxygen * molar volume of oxygen)

Step 5: The calculated volume of oxygen is the same as the volume of air required for combustion, as air contains approximately 21% oxygen.

Unfortunately, the molar volumes of the components and the molar volume of oxygen are not provided in the given information. Therefore, we cannot perform the calculations without this crucial information.