A mixture of 16 g `CH_(4)` and 64 g of `O_(2)` is ignited in a sealed bulb of 3L and then cooled to 27°C. The pressure in the bulb will be

Options

A. 0.82 atm
B. 8.2 atm
C. 24.6 atm
D. 2.46 atm

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of methane (CH₄) with oxygen (O₂).

The balanced equation for the combustion of methane is:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]

### Step 2: Calculate the number of moles of CH₄ and O₂.

- Molar mass of CH₄ (methane) = 12 (C) + 4 (H) = 16 g/mol
- Molar mass of O₂ (oxygen) = 16 × 2 = 32 g/mol

Now, calculate the moles:
- Moles of CH₄:
\[ n_{\text{CH}_4} = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mole} \]

- Moles of O₂:
\[ n_{\text{O}_2} = \frac{64 \text{ g}}{32 \text{ g/mol}} = 2 \text{ moles} \]

### Step 3: Determine the limiting reactant.

From the balanced equation, 1 mole of CH₄ reacts with 2 moles of O₂. Since we have 1 mole of CH₄ and 2 moles of O₂, CH₄ is the limiting reactant.

### Step 4: Calculate the moles of products formed.

From the balanced equation:
- 1 mole of CH₄ produces 1 mole of CO₂ and 2 moles of H₂O.
Thus, after the reaction:
- Moles of CO₂ = 1 mole
- Moles of H₂O = 2 moles

Total moles of gas after the reaction:
\[ n_{\text{total}} = n_{\text{CO}_2} + n_{\text{H}_2\text{O}} = 1 + 2 = 3 \text{ moles} \]

### Step 5: Use the ideal gas law to find the pressure.

The ideal gas law is given by:
\[ PV = nRT \]

Where:
- \( P \) = pressure (atm)
- \( V \) = volume (L)
- \( n \) = number of moles
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature (K)

Convert the temperature from Celsius to Kelvin:
\[ T = 27 + 273 = 300 \text{ K} \]

Now, substituting the values into the ideal gas law:
\[ P = \frac{nRT}{V} \]
\[ P = \frac{3 \text{ moles} \times 0.0821 \text{ L·atm/(K·mol)} \times 300 \text{ K}}{3 \text{ L}} \]

Calculating:
\[ P = \frac{3 \times 0.0821 \times 300}{3} \]
\[ P = 0.0821 \times 300 \]
\[ P = 24.63 \text{ atm} \]

### Step 6: Final pressure calculation.

However, since the total pressure is calculated based on the moles of gas produced, we need to adjust for the fact that the volume is 3L:
\[ P = \frac{24.63}{3} = 8.21 \text{ atm} \]

### Final Answer:
The pressure in the bulb will be approximately **8.2 atm**.

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