The % by volume of `C_(4)H_(10)` in a gaseous mixture of `C_(4)H_(10)` and `CO` is 40. When `200ml` of the mixture is burnt in excess of `O_(2)`. Find volume (in ml) of `CO_(2)` produced.

Options

A. `220`
B. `340`
C. `440`
D. `560`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the volume of `C4H10` in the mixture.
Given that the percentage by volume of `C4H10` in the mixture is 40%, we can calculate the volume of `C4H10` in the 200 ml mixture.

\[ \text{Volume of } C4H10 = \frac{40}{100} \times 200 \text{ ml} = 80 \text{ ml} \]

### Step 2: Determine the volume of `CO` in the mixture.
Since the total volume of the mixture is 200 ml, we can find the volume of `CO` by subtracting the volume of `C4H10` from the total volume.

\[ \text{Volume of } CO = 200 \text{ ml} - 80 \text{ ml} = 120 \text{ ml} \]

### Step 3: Write the balanced chemical equations for combustion.
The combustion reactions for `C4H10` and `CO` are as follows:

1. For `C4H10`:
\[ C4H10 + 6O2 \rightarrow 4CO2 + 5H2O \]
From this equation, 1 volume of `C4H10` produces 4 volumes of `CO2`.

2. For `CO`:
\[ 2CO + O2 \rightarrow 2CO2 \]
From this equation, 1 volume of `CO` produces 1 volume of `CO2`.

### Step 4: Calculate the volume of `CO2` produced from `C4H10`.
Using the volume of `C4H10` we calculated earlier (80 ml), we can find the volume of `CO2` produced.

\[ \text{Volume of } CO2 \text{ from } C4H10 = 80 \text{ ml} \times 4 = 320 \text{ ml} \]

### Step 5: Calculate the volume of `CO2` produced from `CO`.
Using the volume of `CO` we calculated earlier (120 ml), we can find the volume of `CO2` produced.

\[ \text{Volume of } CO2 \text{ from } CO = 120 \text{ ml} \times 1 = 120 \text{ ml} \]

### Step 6: Calculate the total volume of `CO2` produced.
Now, we can find the total volume of `CO2` produced from both `C4H10` and `CO`.

\[ \text{Total Volume of } CO2 = 320 \text{ ml} + 120 \text{ ml} = 440 \text{ ml} \]

### Final Answer:
The total volume of `CO2` produced when 200 ml of the mixture is burnt in excess of `O2` is **440 ml**.

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