`200ml` of a gaseous mixture containing `CO,CO_(2)` and `N_(2)` on complete combusion in just sufficient amount of `O_(2)` showed contration of `40ml`. When the resulting gases were passed through `KOH` solution it reduces by `50%` then calculate the volume ratio of `V_(co_(2)): V_(CO): V_(N_(2))` in original mixture .

Options

A. `4:1:5`
B. `2:3:5`
C. `1:4:5`
D. `1:3:5`

Explanation

To solve the problem, we need to analyze the information given about the gaseous mixture and the reactions that occur during combustion and absorption in KOH solution.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
We have a gaseous mixture of `CO`, `CO2`, and `N2` in a volume of `200 ml`. Upon complete combustion with sufficient `O2`, the volume of gas contracts to `40 ml`.

2. **Calculating the Volume of Gases Consumed:**
The contraction in volume is given as:
\[ \text{Initial Volume} - \text{Final Volume} = 200 ml - 40 ml = 160 ml \]
This 160 ml represents the volume of gases that reacted (i.e., `CO` and `CO2`).

3. **Identifying the Gases Involved in Combustion:**
The combustion reactions are as follows:
- For `CO`:
\[ 2CO + O2 \rightarrow 2CO2 \]
- For `CO2`:
\[ CO2 + O2 \rightarrow \text{(no reaction, it does not combust)} \]
- `N2` does not react in combustion.

4. **Volume of Gases Absorbed by KOH:**
After combustion, the resulting gases are passed through KOH solution, which absorbs `CO2`. The problem states that the volume reduces by `50%` when passed through KOH. Therefore:
\[ \text{Volume after KOH} = 40 ml \times 0.5 = 20 ml \]
This means that `20 ml` of gas remains after `CO2` is absorbed.

5. **Calculating the Volume of `CO2` Absorbed:**
The volume of `CO2` absorbed by KOH is:
\[ \text{Volume of CO2} = 40 ml - 20 ml = 20 ml \]

6. **Setting Up the Volume Relationships:**
Let’s denote:
- Volume of `CO` = \( V_{CO} \)
- Volume of `CO2` = \( V_{CO2} \)
- Volume of `N2` = \( V_{N2} \)

From the combustion reactions, we know:
\[ V_{CO} + V_{CO2} = 160 ml \quad \text{(total volume of gases that reacted)} \]
And we also know:
\[ V_{CO2} = 20 ml \quad \text{(volume of CO2 absorbed)} \]

7. **Substituting the Known Values:**
Substituting \( V_{CO2} \) into the first equation:
\[ V_{CO} + 20 ml = 160 ml \]
\[ V_{CO} = 160 ml - 20 ml = 140 ml \]

8. **Finding the Volume of `N2`:**
The total initial volume of the mixture is `200 ml`, so:
\[ V_{CO} + V_{CO2} + V_{N2} = 200 ml \]
Substituting the known values:
\[ 140 ml + 20 ml + V_{N2} = 200 ml \]
\[ V_{N2} = 200 ml - 160 ml = 40 ml \]

9. **Calculating the Volume Ratio:**
Now we have:
- \( V_{CO2} = 20 ml \)
- \( V_{CO} = 140 ml \)
- \( V_{N2} = 40 ml \)

The volume ratio \( V_{CO2} : V_{CO} : V_{N2} \) is:
\[ V_{CO2} : V_{CO} : V_{N2} = 20 : 140 : 40 \]
Simplifying this ratio:
\[ = 1 : 7 : 2 \]

### Final Answer:
The volume ratio of \( V_{CO2} : V_{CO} : V_{N2} \) in the original mixture is \( 1 : 7 : 2 \).