What is the mass (in grams) of 6.00 L of propane vapor (C:Hs) at STP?

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Question:** What is the mass (in grams) of 6.00 L of propane vapor (C₃H₈) at STP?

**Explanation:**

To find the mass of 6.00 L of propane vapor at STP (Standard Temperature and Pressure), we need to use the ideal gas law and the molar volume of a gas at STP. 

1. **Molar Volume at STP:** At STP, 1 mole of any ideal gas occupies 22.4 L.
2. **Molar Mass of Propane (C₃H₈):** Calculate the molar mass using atomic masses:
   - Carbon (C): 12.01 g/mol
   - Hydrogen (H): 1.01 g/mol
   - Molar mass of C₃H₈ = (3 × 12.01) + (8 × 1.01) = 44.09 g/mol

3. **Volume-Mole Relationship:** 
   \[
   \text{Moles of propane} = \frac{6.00 \, \text{L}}{22.4 \, \text{L/mol}}
   \]

4. **Calculate Mass:**
   - Mass = Moles × Molar Mass
   - Use the moles obtained from the volume-mole relationship to find the mass.

This calculation provides the mass of 6.00 L of propane vapor at STP.
Transcribed Image Text:**Question:** What is the mass (in grams) of 6.00 L of propane vapor (C₃H₈) at STP? **Explanation:** To find the mass of 6.00 L of propane vapor at STP (Standard Temperature and Pressure), we need to use the ideal gas law and the molar volume of a gas at STP. 1. **Molar Volume at STP:** At STP, 1 mole of any ideal gas occupies 22.4 L. 2. **Molar Mass of Propane (C₃H₈):** Calculate the molar mass using atomic masses: - Carbon (C): 12.01 g/mol - Hydrogen (H): 1.01 g/mol - Molar mass of C₃H₈ = (3 × 12.01) + (8 × 1.01) = 44.09 g/mol 3. **Volume-Mole Relationship:** \[ \text{Moles of propane} = \frac{6.00 \, \text{L}}{22.4 \, \text{L/mol}} \] 4. **Calculate Mass:** - Mass = Moles × Molar Mass - Use the moles obtained from the volume-mole relationship to find the mass. This calculation provides the mass of 6.00 L of propane vapor at STP.
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