Butane `(C_(4)H_(10))` gas burns in oxygen to give carbon dioxide and water accoding 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 ini excess of oxygen at `67^(@)C` and 2 bar pressure, calculate the dioxide evolved.

Explanation

To solve the problem of how much carbon dioxide is evolved when 5.0 L of butane is burned in excess oxygen at 67°C and 2 bar pressure, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the combustion of butane is:
\[ 2C_{4}H_{10} (g) + 13O_{2}(g) \rightarrow 8 CO_{2}(g) + 10H_{2}O(l) \]

### Step 2: Determine the volume ratio
From the balanced equation, we see that 2 volumes of butane produce 8 volumes of carbon dioxide. Thus, the volume ratio of butane to carbon dioxide is:
\[ \frac{8 \text{ L CO}_2}{2 \text{ L C}_4H_{10}} = 4 \]

### Step 3: Calculate the volume of carbon dioxide produced
If 5.0 L of butane is burned, the volume of carbon dioxide produced can be calculated using the ratio:
\[ \text{Volume of } CO_2 = 5.0 \, L \times \frac{8 \, L \, CO_2}{2 \, L \, C_4H_{10}} = 5.0 \, L \times 4 = 20.0 \, L \]

### Step 4: Adjust for the conditions of temperature and pressure
To find the volume of carbon dioxide at the given conditions (67°C and 2 bar), we will use the ideal gas law and the concept of gas volume under different conditions.

### Step 5: Convert temperature to Kelvin
Convert the temperature from Celsius to Kelvin:
\[ T = 67 + 273 = 340 \, K \]

### Step 6: Use the ideal gas law to find the volume at standard conditions
Using the ideal gas law, we can relate the volumes at different conditions:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_1 = 1.013 \, \text{bar} \) (standard pressure)
- \( V_1 = 20.0 \, L \) (volume of \( CO_2 \) at standard conditions)
- \( T_1 = 273 \, K \) (standard temperature)
- \( P_2 = 2 \, \text{bar} \) (given pressure)
- \( T_2 = 340 \, K \) (given temperature)
- \( V_2 \) is the volume of \( CO_2 \) at the given conditions.

### Step 7: Rearranging the ideal gas law to solve for \( V_2 \)
Rearranging gives:
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

### Step 8: Plug in the values
Substituting the values into the equation:
\[ V_2 = \frac{(1.013 \, \text{bar}) (20.0 \, L) (340 \, K)}{(2 \, \text{bar}) (273 \, K)} \]

### Step 9: Calculate \( V_2 \)
Calculating the above expression:
\[ V_2 = \frac{1.013 \times 20.0 \times 340}{2 \times 273} \]
\[ V_2 \approx \frac{6862.2}{546} \approx 12.58 \, L \]

Thus, the volume of carbon dioxide evolved when 5.0 L of butane is burned in excess oxygen at 67°C and 2 bar pressure is approximately **12.58 L**.