Math Problem Statement

A 10.00 L tank at -1.6 °C is filled with 13.7 g of carbon monoxide gas and 7.95 g of dinitrogen monoxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction of each gas. Be sure each of your answer entries has the correct number of significant digits.

Solution

To calculate the mole fraction of each gas, we need to determine the moles of both carbon monoxide (CO) and dinitrogen monoxide (N₂O) first, and then use these values to find their respective mole fractions.

Given Information

  • Volume of tank (V): 10.00 L
  • Temperature (T): -1.6 °C = 271.55 K (by converting to Kelvin with T(K)=T(°C)+273.15T(K) = T(°C) + 273.15)
  • Mass of carbon monoxide (CO): 13.7 g
  • Mass of dinitrogen monoxide (N₂O): 7.95 g
  • Assume ideal gas behavior.

Molar Mass of Each Gas

  1. Carbon monoxide (CO): 12.01g/mol+16.00g/mol=28.01g/mol12.01 \, \text{g/mol} + 16.00 \, \text{g/mol} = 28.01 \, \text{g/mol}
  2. Dinitrogen monoxide (N₂O): 2×14.01g/mol+16.00g/mol=44.02g/mol2 \times 14.01 \, \text{g/mol} + 16.00 \, \text{g/mol} = 44.02 \, \text{g/mol}

Moles of Each Gas

The formula for moles is: n=massmolar massn = \frac{\text{mass}}{\text{molar mass}}

  1. Moles of CO: nCO=13.7g28.01g/mol0.489moln_{\text{CO}} = \frac{13.7 \, \text{g}}{28.01 \, \text{g/mol}} \approx 0.489 \, \text{mol}

  2. Moles of N₂O: nN₂O=7.95g44.02g/mol0.181moln_{\text{N₂O}} = \frac{7.95 \, \text{g}}{44.02 \, \text{g/mol}} \approx 0.181 \, \text{mol}

Total Moles

ntotal=nCO+nN₂O=0.489+0.181=0.670moln_{\text{total}} = n_{\text{CO}} + n_{\text{N₂O}} = 0.489 + 0.181 = 0.670 \, \text{mol}

Mole Fraction of Each Gas

The mole fraction, XX, is calculated as: XCO=nCOntotalX_{\text{CO}} = \frac{n_{\text{CO}}}{n_{\text{total}}} XN₂O=nN₂OntotalX_{\text{N₂O}} = \frac{n_{\text{N₂O}}}{n_{\text{total}}}

  1. Mole fraction of CO: XCO=0.4890.6700.730X_{\text{CO}} = \frac{0.489}{0.670} \approx 0.730

  2. Mole fraction of N₂O: XN₂O=0.1810.6700.270X_{\text{N₂O}} = \frac{0.181}{0.670} \approx 0.270

Final Answer

  • Mole fraction of CO: 0.730
  • Mole fraction of N₂O: 0.270

Would you like further clarification on any step?


Further Questions to Explore

  1. How would the mole fractions change if the temperature were increased?
  2. What would happen to the mole fractions if more CO were added to the tank?
  3. How is the mole fraction related to the partial pressure in a gas mixture?
  4. What is the significance of using ideal gas behavior in this calculation?
  5. How would you convert the mole fraction into a percentage composition of each gas?

Tip:

Always double-check unit conversions, especially for temperature, as they can significantly affect calculations in gas laws and stoichiometry.

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Math Problem Analysis

Mathematical Concepts

Stoichiometry
Ideal Gas Law
Mole Fractions
Molar Mass

Formulas

n = mass / molar mass
X = n_gas / n_total

Theorems

Ideal Gas Law
Mole Fraction Calculation

Suitable Grade Level

Grades 11-12 or College-level Chemistry