1.2g of carbon is burnt completely in oxygen (limted supply) to produce CO and `CO_(2)` . This mixture of gases is treated with solid `I_(2)O_(5)` (to know the amount of CO produced). The librated iodine required 120 ml of 0.1M hypo solution for complete titration. The percentage carbon converted into CO is :

Options

A. 0.6
B. 1
C. 0.5
D. 0.3

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the number of moles of carbon.
We start with the mass of carbon given in the problem:
- Mass of carbon = 1.2 g
- Molar mass of carbon (C) = 12 g/mol

Using the formula for moles:
\[ \text{Number of moles of carbon} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.2 \, \text{g}}{12 \, \text{g/mol}} = 0.1 \, \text{mol} = 100 \, \text{mmol} \]

### Step 2: Calculate the moles of hypo solution used.
The volume of hypo solution used is given as 120 mL with a concentration of 0.1 M. We can find the number of moles of hypo (sodium thiosulfate, Na2S2O3) using the formula:
\[ \text{Number of moles} = \text{Volume (L)} \times \text{Molarity (mol/L)} \]
Converting 120 mL to liters:
\[ \text{Volume} = 120 \, \text{mL} = 0.120 \, \text{L} \]
Now calculate the moles:
\[ \text{Moles of hypo} = 0.120 \, \text{L} \times 0.1 \, \text{mol/L} = 0.012 \, \text{mol} = 12 \, \text{mmol} \]

### Step 3: Determine the moles of iodine produced.
From the reaction between iodine (I2) and hypo, we know:
\[ \text{2 moles of Na2S2O3} \text{ react with } \text{1 mole of I2} \]
Thus, the moles of iodine produced can be calculated as:
\[ \text{Moles of I2} = \frac{12 \, \text{mmol}}{2} = 6 \, \text{mmol} \]

### Step 4: Relate iodine to carbon monoxide produced.
From the reaction of carbon monoxide (CO) with I2O5:
\[ \text{1 mole of I2} \text{ reacts with } 5 \text{ moles of CO} \]
Thus, the moles of CO produced can be calculated as:
\[ \text{Moles of CO} = 5 \times \text{Moles of I2} = 5 \times 6 \, \text{mmol} = 30 \, \text{mmol} \]

### Step 5: Calculate the percentage of carbon converted into CO.
Now we know:
- Initial moles of carbon = 100 mmol
- Moles of carbon converted into CO = 30 mmol

The percentage of carbon converted into CO is given by:
\[ \text{Percentage of carbon converted into CO} = \left( \frac{\text{moles of CO}}{\text{initial moles of carbon}} \right) \times 100 = \left( \frac{30 \, \text{mmol}}{100 \, \text{mmol}} \right) \times 100 = 30\% \]

### Final Answer:
The percentage of carbon converted into CO is **30%**.

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