2 grams of a gas mixture of CO and `CO_2` on reaction with excess `I_2 O_5` yields 2.54 grams of `I_2`. What would be the mass% of CO in the original mixture?

Options

A. `70%`
B. `95.3%`
C. `88.4%`
D. `80.7%`

Answer

To solve the problem step by step, we will follow the procedure outlined in the video transcript while providing a clear explanation for each step.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction of carbon monoxide (CO) with iodine pentoxide (I₂O₅) can be represented as:
\[ 5 \text{CO} + \text{I}_2\text{O}_5 \rightarrow 5 \text{CO}_2 + \text{I}_2 \]
This indicates that 5 moles of CO produce 1 mole of I₂.

2. **Calculate Moles of I₂ Produced**:
We are given that 2.54 grams of I₂ are produced. To find the number of moles of I₂, we use the molar mass of I₂, which is approximately 254 g/mol.
\[ \text{Moles of I}_2 = \frac{\text{mass of I}_2}{\text{molar mass of I}_2} = \frac{2.54 \text{ g}}{254 \text{ g/mol}} = 0.01 \text{ moles} \]

3. **Determine Moles of CO Required**:
According to the stoichiometry of the reaction, 1 mole of I₂ is produced from 5 moles of CO. Therefore, the moles of CO required can be calculated as:
\[ \text{Moles of CO} = 5 \times \text{Moles of I}_2 = 5 \times 0.01 = 0.05 \text{ moles} \]

4. **Calculate Mass of CO**:
The molar mass of CO is approximately 28 g/mol (12 g/mol for C and 16 g/mol for O). Thus, the mass of CO can be calculated as:
\[ \text{Mass of CO} = \text{Moles of CO} \times \text{Molar mass of CO} = 0.05 \text{ moles} \times 28 \text{ g/mol} = 1.40 \text{ g} \]

5. **Calculate Mass Percent of CO in the Mixture**:
The total mass of the gas mixture is given as 2 grams. The mass percent of CO in the original mixture can be calculated using the formula:
\[ \text{Mass \% of CO} = \left( \frac{\text{Mass of CO}}{\text{Total mass of mixture}} \right) \times 100 = \left( \frac{1.40 \text{ g}}{2 \text{ g}} \right) \times 100 = 70\% \]

### Final Answer:
The mass percent of CO in the original mixture is **70%**.