`50mL` of `CO` is mixed with `20mL` of oxygen and sparked. After the reaction, the mixture is treated with an aqueous `KOH` solution. Choose the correct option :

Options

A. the volume of `CO` that reacts `= 30 mL`
B. volume of `CO_(2)` formed `=50mL`
C. volume of `CO` that remains after treatment with `KOH = 10mL`
D. the volume of the `CO` that remains after treatment with `KOH = 20mL`

Answer

To solve the problem step by step, we will analyze the reaction between carbon monoxide (CO) and oxygen (O2), and then determine the amount of CO that remains after treatment with aqueous KOH.

### Step 1: Write the balanced chemical equation
The reaction between carbon monoxide and oxygen can be represented by the following balanced equation:
\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]
This indicates that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2.

### Step 2: Determine the initial volumes of reactants
We have:
- Volume of CO = 50 mL
- Volume of O2 = 20 mL

### Step 3: Determine the stoichiometric requirements
From the balanced equation, we see that 1 mole of O2 reacts with 2 moles of CO. Therefore, for 20 mL of O2, the amount of CO required is:
\[ \text{Required CO} = 2 \times \text{Volume of O2} = 2 \times 20 \text{ mL} = 40 \text{ mL} \]

### Step 4: Identify the limiting reactant
We have 50 mL of CO available, but only 40 mL is required to react with 20 mL of O2. Therefore, O2 is the limiting reactant in this reaction.

### Step 5: Calculate the volume of products formed
Since 20 mL of O2 reacts with 40 mL of CO, the reaction will produce:
- Volume of CO2 produced = 20 mL O2 will produce 20 mL CO2 (since 2 moles of CO produce 2 moles of CO2).

### Step 6: Determine the remaining volume of CO
After the reaction, the remaining volume of CO can be calculated as follows:
- Initial volume of CO = 50 mL
- Volume of CO reacted = 40 mL (since 20 mL of O2 reacts with 40 mL of CO)
- Remaining volume of CO = 50 mL - 40 mL = 10 mL

### Step 7: Treatment with KOH
When the remaining gas mixture (which contains CO and CO2) is treated with aqueous KOH, the KOH will absorb CO2, as KOH reacts with CO2 to form potassium carbonate (K2CO3). Therefore, the CO2 will be removed from the mixture.

### Step 8: Final volume of CO after treatment with KOH
After treating the mixture with KOH:
- Volume of CO2 absorbed = 20 mL (produced from the reaction)
- Volume of CO remaining = 10 mL (calculated previously)

Thus, the final volume of CO after treatment with KOH is:
\[ \text{Final volume of CO} = 10 \text{ mL} \]

### Conclusion
The correct option is that the volume of CO that remains after treating with KOH is **10 mL**.

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