A mixture of ethane `(C_(2)H_(6))` and ethene `(C_(2)H_(4))` occupies `40 L` at `1.00 atm` and at `400 K`. The mixture reacts completely with `130 g` of `O_(2)` to produce `CO_(2)` and `H_(2)O`. Assuming ideal gas behaviour, calculate the mole fractions of `C_(2)H_(4)` and `C_(2)H_(6)` in the mixture.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of \( O_2 \)

Using the ideal gas law equation \( PV = nRT \):

- \( P = 1.00 \, \text{atm} \)
- \( V = 40 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 400 \, \text{K} \)

Rearranging the equation to find \( n \):

\[ n = \frac{PV}{RT} \]

Substituting the values:

\[ n = \frac{(1.00 \, \text{atm}) \times (40 \, \text{L})}{(0.0821 \, \text{L atm/(K mol)}) \times (400 \, \text{K})} \]

Calculating:

\[ n = \frac{40}{32.84} \approx 1.218 \, \text{moles of } O_2 \]

### Step 2: Write the balanced chemical equations

1. For ethane \( (C_2H_6) \):
\[ C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O \]

2. For ethene \( (C_2H_4) \):
\[ C_2H_4 + 3 O_2 \rightarrow 2 CO_2 + 2 H_2O \]

### Step 3: Define variables for moles of each component

Let:
- \( A \) = moles of \( C_2H_4 \)
- \( B \) = moles of \( C_2H_6 \)

From the total moles of the gas mixture:

\[ A + B = 1.218 \]

### Step 4: Set up the equation based on oxygen consumption

From the reactions, the moles of \( O_2 \) consumed can be expressed as:

\[ \frac{7}{2}A + 3B = 1.218 \]

### Step 5: Solve the equations

Substituting \( B = 1.218 - A \) into the oxygen consumption equation:

\[ \frac{7}{2}A + 3(1.218 - A) = 1.218 \]

Expanding and simplifying:

\[ \frac{7}{2}A + 3.654 - 3A = 1.218 \]

Combining like terms:

\[ \left(\frac{7}{2} - 3\right)A = 1.218 - 3.654 \]
\[ \left(\frac{1}{2}\right)A = -2.436 \]
\[ A = 0.817 \, \text{moles of } C_2H_4 \]

Now substituting back to find \( B \):

\[ B = 1.218 - 0.817 = 0.401 \, \text{moles of } C_2H_6 \]

### Step 6: Calculate mole fractions

Mole fraction of \( C_2H_4 \):

\[ X_{C_2H_4} = \frac{A}{A + B} = \frac{0.817}{1.218} \approx 0.6707 \]

Mole fraction of \( C_2H_6 \):

\[ X_{C_2H_6} = \frac{B}{A + B} = \frac{0.401}{1.218} \approx 0.3293 \]

### Final Answer:

- Mole fraction of \( C_2H_4 \approx 0.6707 \)
- Mole fraction of \( C_2H_6 \approx 0.3293 \)

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