Problem 3
Question
A gas is a mixture of \(22 \% \mathrm{O}_{2}, 33 \% \mathrm{~N}_{2}\), and \(45 \% \mathrm{CO}_{2}\) by volume. Calculate (a) the mole fraction of the constituents in the mixture (b) the mixture molecular weight \(\mathrm{MW}_{\mathrm{m}}\)
Step-by-Step Solution
Verified Answer
The mole fractions of \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{CO}_{2}\) are 0.22, 0.33, and 0.45 respectively. And the mixture molecular weight, \(\mathrm{MW}_{\mathrm{m}}\), is \(32 \times 0.22 + 28 \times 0.33 + 44 \times 0.45\) g/mol.
1Step 1: Mole Fraction Calculation
For a gas, under the same physical conditions, the volume percentage is identical to the mole fraction. Therefore, to calculate the mole fraction, simply convert the given volume percentages into decimal form. For \(\mathrm{O}_{2}\), 22 percent becomes 0.22. For \(\mathrm{N}_{2}\), 33 percent becomes 0.33. For \(\mathrm{CO}_{2}\), 45 percent becomes 0.45.
2Step 2: Molecular Weight of Individual Gases
Determining the individual molecular weights of the gases. This can be done using a periodic table or similar resource. For \(\mathrm{O}_{2}\), the double amount of Oxygen atoms will be \(16 \times 2 = 32 \mathrm{~g/mol}\). For \(\mathrm{N}_{2}\), this will be \(14 \times 2 = 28 \mathrm{~g/mol}\). And for \(\mathrm{CO}_{2}\), it is one Carbon atom and 2 Oxygen atoms, hence, \(12 + 16 \times 2 = 44 \mathrm{~g/mol}\).
3Step 3: Mixture Molecular Weight Calculation
The Mixture Molecular Weight, \(\mathrm{MW}_{\mathrm{m}}\), is found by the sum of the product of each individual molecular weight and its corresponding mole fraction. Therefore, \(\mathrm{MW}_{\mathrm{m}} = \mathrm{MW}_{\mathrm{O}_{2}} \times X_{\mathrm{O}_{2}} + \mathrm{MW}_{\mathrm{N}_{2}} \times X_{\mathrm{N}_{2}} + \mathrm{MW}_{\mathrm{CO}_{2}} \times X_{\mathrm{CO}_{2}} = 32 \times 0.22 + 28 \times 0.33 + 44 \times 0.45\)
Key Concepts
Mole FractionGas MixtureMolecular Weight of Gases
Mole Fraction
Mole Fraction is a way to express the concentration of a component in a mixture. In the context of gases, it helps us understand the proportion of each individual gas in the mix. The mole fraction (\[ X_i \]) of a component i is calculated by taking the moles of the component and dividing it by the total moles in the mixture. However, for gases under the same conditions of temperature and pressure, the volume percentages provided directly translate to mole fractions due to the ideal gas law.
- To translate a percentage to a mole fraction, you divide the percent by 100. For example, if the volume percentage of Oxygen (\(\mathrm{O}_2\)) in a gas mixture is 22%, its mole fraction will be 0.22.
- The same applies to Nitrogen (\(\mathrm{N}_2\)) and Carbon Dioxide (\(\mathrm{CO}_2\)), with mole fractions of 0.33 and 0.45, respectively.
Gas Mixture
A Gas Mixture is comprised of more than one component, allowing us to model the behavior of real gases in different environments. It is significant in fields like chemistry, environmental science, and engineering. When combining gases, each retains its properties within the mixture, and the overall behavior is determined by the blend of these components.
An important concept when dealing with gas mixtures is the partial pressure of each gas. Each gas in a mixture exerts pressure independently, and these individual pressures sum up to the total pressure of the mixture.
- Each gas contributes to the total pressure in proportion to its mole fraction.
- The behavior of gases in a mixture is often predicted using Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of its components.
Molecular Weight of Gases
The Molecular Weight of Gases is a critical concept for understanding the mass and composition of a gas. The molecular weight (MW) of a gas is calculated by summing the atomic weights of all atoms present in a molecule and is expressed in grams per mole (g/mol).To determine the molecular weight of a gas mixture, calculate the average based on individual molecular weights and mole fractions of each gas:
- First, find the molecular weight of each gas. For example, \( \mathrm{O}_2 = 32 \text{ g/mol} \), \( \mathrm{N}_2 = 28 \text{ g/mol} \), and \( \mathrm{CO}_2 = 44 \text{ g/mol} \).
- Then, use the formula: \[ \mathrm{MW}_m = (\mathrm{MW}_{\mathrm{O}_2} \times X_{\mathrm{O}_2}) + (\mathrm{MW}_{\mathrm{N}_2} \times X_{\mathrm{N}_2}) + (\mathrm{MW}_{\mathrm{CO}_2} \times X_{\mathrm{CO}_2}) \]
- Substitute the values: \[ \mathrm{MW}_m = 32 \times 0.22 + 28 \times 0.33 + 44 \times 0.45 \]
Other exercises in this chapter
Problem 1
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Write the chemical reaction for the complete combustion of JP-4 and air. JP-4 has the formula \(\mathrm{CH}_{1.93}\). Also, calculate the stoichiometric fuel- t
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