Forty millililtre of `CO` was mixed with 100 ml of `O_2` and the mixture was exploded. On cooling, the reaction mixture was shaken with `KOH` What volume of gas is left?

Options

A. 60 ml of `O_2`
B. 80 ml of `O_2`
C. 20 ml of `CO`
D. 40 ml `CO_2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical reaction
The reaction between carbon monoxide (CO) and oxygen (O2) can be represented as follows:
\[ \text{CO} + \frac{1}{2} \text{O}_2 \rightarrow \text{CO}_2 \]

### Step 2: Determine the volumes of reactants
We are given:
- Volume of CO = 40 ml
- Volume of O2 = 100 ml

### Step 3: Calculate the amount of O2 required for the reaction
From the balanced equation, we see that 1 ml of CO reacts with 0.5 ml of O2. Therefore, for 40 ml of CO:
\[ \text{O}_2 \text{ required} = 40 \text{ ml CO} \times \frac{0.5 \text{ ml O}_2}{1 \text{ ml CO}} = 20 \text{ ml O}_2 \]

### Step 4: Determine the amount of O2 left after the reaction
Initially, we had 100 ml of O2. After the reaction, 20 ml of O2 is consumed:
\[ \text{O}_2 \text{ left} = 100 \text{ ml} - 20 \text{ ml} = 80 \text{ ml} \]

### Step 5: Identify the product formed
The reaction produces CO2. The volume of CO2 produced will be equal to the volume of CO reacted, which is 40 ml.

### Step 6: Consider the effect of KOH
KOH is used to absorb CO2. Since all the CO2 produced (40 ml) will be absorbed by KOH, it does not contribute to the volume of gas left.

### Final Calculation
After the reaction and absorption of CO2 by KOH, the only gas left is the unreacted O2:
- Volume of gas left = Volume of O2 left = 80 ml

### Conclusion
The volume of gas left after the reaction and treatment with KOH is **80 ml**.

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