One litre sample of a mixture of `CH_(4)` and `O_(2)` measured at `32^(@)C` and `760` torr, was allowed to react at constant pressure in a calorimeter which together with its content has a heat capacity of 1260 cal/degree. The complete combustion of `CH_(4)` to `CO_(2)` and water caused a temperature rise in calorimeter of `0.667K`. calculate mole `%` of `CH_(4)` in original mixture.
[Given: Heat of combustion of `CH_(4)` is `-210.8 Kcal//mol`. Total heat capacity of the calorimeter `= 2108 cal K`]

Explanation

To solve the problem step by step, we will follow the outlined procedure:

### Step 1: Calculate the heat generated (Q) during the combustion of CH₄

The heat generated can be calculated using the formula:

\[ Q = C \times \Delta T \]

Where:
- \( C \) is the heat capacity of the calorimeter (1260 cal/°C)
- \( \Delta T \) is the change in temperature (0.667 K)

Substituting the values:

\[ Q = 1260 \, \text{cal/°C} \times 0.667 \, \text{K} = 840.42 \, \text{cal} \]

### Step 2: Convert the heat generated (Q) into kilocalories

Since the heat of combustion is given in kilocalories, we need to convert the heat generated from calories to kilocalories:

\[ Q = \frac{840.42 \, \text{cal}}{1000} = 0.84042 \, \text{kcal} \]

### Step 3: Calculate the moles of CH₄ that reacted

Using the heat of combustion of CH₄, which is given as -210.8 kcal/mol, we can calculate the moles of CH₄ that reacted:

\[ \text{Moles of } CH₄ = \frac{Q}{\text{Heat of combustion of } CH₄} \]

Substituting the values:

\[ \text{Moles of } CH₄ = \frac{0.84042 \, \text{kcal}}{210.8 \, \text{kcal/mol}} \approx 0.00398 \, \text{mol} \]

### Step 4: Calculate the total moles of the gas mixture

Using the ideal gas law, we can calculate the total moles of the gas mixture:

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

Where:
- \( P = 760 \, \text{torr} = 1 \, \text{atm} \) (since 760 torr = 1 atm)
- \( V = 1 \, \text{L} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 32 \, \text{°C} = 305 \, \text{K} \) (convert to Kelvin)

Substituting the values:

\[ n = \frac{(1 \, \text{atm}) \times (1 \, \text{L})}{(0.0821 \, \text{L atm/(K mol)}) \times (305 \, \text{K})} \approx 0.040 \, \text{mol} \]

### Step 5: Calculate the mole percentage of CH₄ in the original mixture

Now we can calculate the mole percentage of CH₄ in the original mixture:

\[ \text{Mole \% of } CH₄ = \left( \frac{\text{Moles of } CH₄}{\text{Total moles}} \right) \times 100 \]

Substituting the values:

\[ \text{Mole \% of } CH₄ = \left( \frac{0.00398 \, \text{mol}}{0.040 \, \text{mol}} \right) \times 100 \approx 9.95\% \]

### Final Answer

The mole percentage of CH₄ in the original mixture is approximately **10%**.

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