Butane `(C_(4)H_(10))` gas burns in oxygen to give carbon dioxide and water according to the reaction.
`2C_(4)H_(10) (g) + 13O_(2)(g) rarr 8 CO_(2)(g) + 10H_(2)O(l)`
When `5.0 L` of butane ware burnt in excess of oxygen at `67^(@)C` and 2 bar pressure, calculate the dioxide evolved.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the combustion of butane `(C₄H₁₀)` in oxygen is:
\[ 2C₄H₁₀(g) + 13O₂(g) \rightarrow 8CO₂(g) + 10H₂O(l) \]

### Step 2: Determine the moles of butane burned
Given that 5.0 L of butane is burned, we need to calculate the volume of carbon dioxide produced based on the stoichiometry of the reaction.

### Step 3: Use stoichiometry to find the volume of CO₂ produced
From the balanced equation, we see that:
- 2 volumes of butane produce 8 volumes of carbon dioxide.

Using the ratio:
\[ \text{Volume of } CO₂ = \left( \frac{8 \text{ L } CO₂}{2 \text{ L } C₄H₁₀} \right) \times 5.0 \text{ L } C₄H₁₀ \]
Calculating this gives:
\[ \text{Volume of } CO₂ = \frac{8}{2} \times 5.0 = 20 \text{ L} \]

### Step 4: Apply the Ideal Gas Law to find the volume of CO₂ under experimental conditions
Now we need to find the volume of CO₂ at the given experimental conditions: 67°C and 2 bar pressure.

First, convert the temperature from Celsius to Kelvin:
\[ T_2 = 67 + 273 = 340 \text{ K} \]

We will use the Ideal Gas Law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_1 = 1.013 \text{ bar} \) (normal pressure)
- \( V_1 = 20 \text{ L} \) (volume of CO₂ at normal conditions)
- \( T_1 = 273 \text{ K} \) (normal temperature)
- \( P_2 = 2 \text{ bar} \) (given pressure)
- \( V_2 \) = volume of CO₂ at experimental conditions
- \( T_2 = 340 \text{ K} \) (given temperature)

### Step 5: Rearranging the equation to solve for \( V_2 \)
Rearranging gives:
\[ V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} \]

### Step 6: Substitute the values into the equation
Substituting the known values:
\[ V_2 = \frac{(1.013 \text{ bar}) \times (20 \text{ L}) \times (340 \text{ K})}{(273 \text{ K}) \times (2 \text{ bar})} \]

### Step 7: Calculate \( V_2 \)
Calculating this gives:
\[ V_2 = \frac{1.013 \times 20 \times 340}{273 \times 2} \approx 12.62 \text{ L} \]

### Final Answer
The volume of carbon dioxide evolved is approximately **12.62 L**.

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