The percentage by volume of `C_(3)H_(8)` in a gaseous mixture of `C_(3)H_(8),CH_(4)` and `CO` is 20. When `100mL` of the mixture is burnt in excess of `O_(2)` the volume of `CO_(2)` produced is :

Options

A. `90mL`
B. `160mL`
C. `140 mL`
D. None of these

Explanation

To solve the problem step by step, we will follow the given information and apply stoichiometry to find the total volume of carbon dioxide produced when the gaseous mixture is burnt.

### Step 1: Determine the volume of each component in the mixture.
The percentage by volume of `C3H8` (propane) in the mixture is given as 20%. Since the total volume of the mixture is 100 mL, we can calculate the volume of `C3H8`:

\[ \text{Volume of } C3H8 = \frac{20}{100} \times 100 \text{ mL} = 20 \text{ mL} \]

The remaining volume, which consists of `CH4` (methane) and `CO` (carbon monoxide), is:

\[ \text{Volume of } CH4 + CO = 100 \text{ mL} - 20 \text{ mL} = 80 \text{ mL} \]

### Step 2: Write the combustion reactions for each component.
1. For `C3H8` (propane):
\[ C3H8 + 5O2 \rightarrow 3CO2 + 4H2O \]
From this reaction, 1 volume of `C3H8` produces 3 volumes of `CO2`.

2. For `CH4` (methane):
\[ CH4 + 2O2 \rightarrow CO2 + 2H2O \]
From this reaction, 1 volume of `CH4` produces 1 volume of `CO2`.

3. For `CO` (carbon monoxide):
\[ 2CO + O2 \rightarrow 2CO2 \]
From this reaction, 1 volume of `CO` produces 1 volume of `CO2`.

### Step 3: Calculate the volume of `CO2` produced from `C3H8`.
Since we have 20 mL of `C3H8`, the volume of `CO2` produced from it is:

\[ \text{Volume of } CO2 \text{ from } C3H8 = 20 \text{ mL} \times 3 = 60 \text{ mL} \]

### Step 4: Calculate the volume of `CO2` produced from `CH4` and `CO`.
Since the total volume of `CH4` and `CO` is 80 mL, and both produce `CO2` in a 1:1 ratio, we can say:

\[ \text{Volume of } CO2 \text{ from } CH4 + CO = 80 \text{ mL} \]

### Step 5: Calculate the total volume of `CO2` produced.
Now, we can find the total volume of `CO2` produced by adding the volumes from `C3H8`, `CH4`, and `CO`:

\[ \text{Total volume of } CO2 = \text{Volume from } C3H8 + \text{Volume from } CH4 + CO \]
\[ \text{Total volume of } CO2 = 60 \text{ mL} + 80 \text{ mL} = 140 \text{ mL} \]

### Final Answer:
The volume of `CO2` produced when 100 mL of the mixture is burnt in excess of `O2` is **140 mL**.

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