A mixture of CO and `CO_(2)` when treated with `I_(2)O_(5)` gives `I_(2)` vapours accoeding to the following equation:
`5CO+I_(2)O_(5)to5CO_(2)+I_(2)`
`I_(2)` vapour was separated and treated with `HClO_(4)` and the resultant `HIO_(4)` required 0.001 " mol of "hlycerol for complete oxidation. After treatment with `I_(2)O_(5)` and removal of `I_(2)` ,t he mixture was treated with excess of 0.1 N NaOH solution and finally this solution required 20 " mL of " 1 N HCl to reach end point using phenolphthalein as indicator, followed by methyl orange as indicator after the first end point, 10 " mL of " further HCl was consumed.
Q. Total number of millimoles of `CO_(2)` in the above problem is

Options

A. 5
B. 10
C. 50
D. 100

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction provided is:
\[ 5 \text{CO} + \text{I}_2\text{O}_5 \rightarrow 5 \text{CO}_2 + \text{I}_2 \]

This means that for every 5 moles of CO that react, 5 moles of CO2 are produced.

### Step 2: Analyze the Reaction with HClO4
The problem states that the iodine vapor (I2) produced requires 0.001 moles of glycerol for complete oxidation to HIO4. This indicates that the amount of I2 produced is related to the amount of CO that reacted.

### Step 3: Determine the Amount of CO
Since 5 moles of CO produce 1 mole of I2, we can find the moles of CO that reacted. Given that 0.001 moles of glycerol are used, we can assume that this corresponds to the amount of I2 produced:
- Moles of I2 produced = 0.001 moles

Using the stoichiometry of the reaction:
\[ 5 \text{CO} \rightarrow 1 \text{I}_2 \]
Thus, for 0.001 moles of I2 produced:
\[ \text{Moles of CO} = 0.001 \times 5 = 0.005 \text{ moles} \]

### Step 4: Convert Moles of CO to Millimoles
To convert moles to millimoles:
\[ 0.005 \text{ moles} = 5 \text{ millimoles of CO} \]

### Step 5: Analyze the NaOH Titration
The mixture was treated with excess 0.1 N NaOH, and it required 20 mL of 1 N HCl to reach the endpoint using phenolphthalein. The reaction of CO2 with NaOH is:
\[ \text{CO}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \]

From the titration:
- 20 mL of 1 N HCl = 20 milliequivalents of HCl
- This HCl neutralizes the NaOH, indicating that the amount of NaOH used is equal to the amount of CO2 present.

### Step 6: Calculate the Total Millimoles of CO2
Since 1 mole of CO2 reacts with 2 equivalents of NaOH:
- The total equivalents of NaOH used = 20 mL × 1 N = 20 equivalents
- Therefore, the moles of CO2 = 20 equivalents / 2 = 10 millimoles of CO2.

### Step 7: Total Millimoles of CO2
We already calculated that there are 5 millimoles of CO and since 5 millimoles of CO produce 5 millimoles of CO2, we add this to the 10 millimoles of CO2 from the titration:
\[ \text{Total millimoles of CO}_2 = 5 + 10 = 15 \text{ millimoles} \]

### Final Answer
The total number of millimoles of CO2 in the above problem is **15 millimoles**.

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