60 g of gaseous `C_(2)H_(6)` are mixed with 28 g of carbon monoxide. The pressure of the resulting gaseous mixture is 3 atm. The partial pressure in atm. Of `C_(2)H_(6)` in the mixture is

Answer

To find the partial pressure of \( C_2H_6 \) in the mixture, we will follow these steps:

### Step 1: Calculate the number of moles of \( C_2H_6 \)

Given:
- Mass of \( C_2H_6 = 60 \, g \)
- Molar mass of \( C_2H_6 = 2 \times 12 + 6 \times 1 = 24 + 6 = 30 \, g/mol \)

Using the formula for moles:
\[ \text{Number of moles of } C_2H_6 = \frac{\text{mass}}{\text{molar mass}} = \frac{60 \, g}{30 \, g/mol} = 2 \, moles \]

### Step 2: Calculate the number of moles of \( CO \)

Given:
- Mass of \( CO = 28 \, g \)
- Molar mass of \( CO = 12 + 16 = 28 \, g/mol \)

Using the formula for moles:
\[ \text{Number of moles of } CO = \frac{28 \, g}{28 \, g/mol} = 1 \, mole \]

### Step 3: Calculate the total number of moles in the mixture

Total number of moles:
\[ \text{Total moles} = \text{moles of } C_2H_6 + \text{moles of } CO = 2 + 1 = 3 \, moles \]

### Step 4: Calculate the mole fraction of \( C_2H_6 \)

Mole fraction of \( C_2H_6 \):
\[ \text{Mole fraction of } C_2H_6 = \frac{\text{moles of } C_2H_6}{\text{total moles}} = \frac{2}{3} \]

### Step 5: Calculate the partial pressure of \( C_2H_6 \)

Given:
- Total pressure of the mixture = 3 atm

Using Dalton's Law of Partial Pressures:
\[ \text{Partial pressure of } C_2H_6 = \text{Mole fraction of } C_2H_6 \times \text{Total pressure} \]
\[ \text{Partial pressure of } C_2H_6 = \frac{2}{3} \times 3 \, atm = 2 \, atm \]

### Final Answer:
The partial pressure of \( C_2H_6 \) in the mixture is **2 atm**.

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