https://file.aiquickdraw.com/mathbot/file/1727548166280ddo4afpn.jpg

To solve the problem, we will follow these steps:

### Given Data:
- Volume of the tank \(V = 10.0 \, \text{L}\)
- Temperature \(T = 3.36^\circ \text{C} = 276.51 \, \text{K}\) (Convert to Kelvin by adding 273.15)
- Mass of chlorine pentafluoride (ClF\(_5\)) \( = 12.4 \, \text{g}\)
- Mass of sulfur hexafluoride (SF\(_6\)) \( = 7.62 \, \text{g}\)
- Molar masses:
  - \(M_{\text{ClF}_5} = 130.443 \, \text{g/mol}\)
  - \(M_{\text{SF}_6} = 146.06 \, \text{g/mol}\)

We assume ideal gas behavior, so we'll use the ideal gas law to calculate the partial pressures:

\[
PV = nRT
\]

Where:
- \(P\) is the pressure (in atm),
- \(V\) is the volume (in liters),
- \(n\) is the number of moles,
- \(R = 0.0821 \, \text{L·atm/mol·K}\) (ideal gas constant),
- \(T\) is the temperature (in Kelvin).

### Step 1: Calculate the number of moles for each gas
\[
n = \frac{\text{mass}}{\text{molar mass}}
\]

#### Chlorine Pentafluoride (ClF\(_5\)):
\[
n_{\text{ClF}_5} = \frac{12.4 \, \text{g}}{130.443 \, \text{g/mol}} 
\]

#### Sulfur Hexafluoride (SF\(_6\)):
\[
n_{\text{SF}_6} = \frac{7.62 \, \text{g}}{146.06 \, \text{g/mol}}
\]

### Step 2: Calculate the total number of moles
\[
n_{\text{total}} = n_{\text{ClF}_5} + n_{\text{SF}_6}
\]

### Step 3: Calculate the mole fraction for each gas
\[
\chi_{\text{ClF}_5} = \frac{n_{\text{ClF}_5}}{n_{\text{total}}}, \quad \chi_{\text{SF}_6} = \frac{n_{\text{SF}_6}}{n_{\text{total}}}
\]

### Step 4: Calculate the partial pressure of each gas using the ideal gas law
\[
P_{\text{ClF}_5} = \chi_{\text{ClF}_5} \times P_{\text{total}}, \quad P_{\text{SF}_6} = \chi_{\text{SF}_6} \times P_{\text{total}}
\]

### Step 5: Calculate the total pressure in the tank using the ideal gas law
\[
P_{\text{total}} = \frac{n_{\text{total}}RT}{V}
\]

Let's proceed with the calculations.### Results:

1. **Moles of chlorine pentafluoride (ClF₅):** 0.0951 mol
2. **Moles of sulfur hexafluoride (SF₆):** 0.0522 mol

3. **Mole fractions:**
   - Chlorine pentafluoride (ClF₅): 0.646
   - Sulfur hexafluoride (SF₆): 0.354

4. **Partial pressures:**
   - Chlorine pentafluoride (ClF₅): 0.216 atm
   - Sulfur hexafluoride (SF₆): 0.118 atm

5. **Total pressure in the tank:** 0.334 atm

Let me know if you need further clarification or additional details!

### 5 Related Questions:
1. How would the total pressure change if the temperature increased?
2. What is the relationship between mole fraction and partial pressure?
3. How does the ideal gas law compare to real gas behavior under these conditions?
4. What would happen to the mole fractions if more sulfur hexafluoride was added?
5. How do you convert between different units of pressure (e.g., atm, Pa)?

### Tip:
Always check units consistency in calculations, especially when using constants like \(R\) in the ideal gas law, to avoid incorrect results.

A 10.0 L tank at 3.36 °C is filled with 12.4 g of chlorine pentafluoride gas and 7.62 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank.

Learn how to calculate the mole fractions and partial pressures of chlorine pentafluoride (ClF5) and sulfur hexafluoride (SF6) in a 10.0 L tank at 3.36°C. Understand key gas laws and the ideal gas law applied in this chemistry problem, perfect for high school and introductory college students.

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