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 outlined process:

### Step 1: Write the Reaction
The reaction between carbon monoxide (CO) and iodine pentoxide (I2O5) can be represented as:
\[ 5 \text{CO} + \text{I}_2\text{O}_5 \rightarrow 5 \text{CO}_2 + \text{I}_2 \]

### Step 2: Calculate the 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 formula:
\[ \text{Number of moles} = \frac{\text{Given weight}}{\text{Molecular mass}} \]

The molecular mass of I2 (iodine) is approximately 254 g/mol. Therefore:
\[ \text{Number of moles of I2} = \frac{2.54 \, \text{g}}{254 \, \text{g/mol}} = 0.01 \, \text{moles} \]

### Step 3: Calculate the Moles of CO Reacted
From the stoichiometry of the reaction, 1 mole of I2 is produced from 5 moles of CO. Thus:
\[ 0.01 \, \text{moles of I2} \rightarrow 5 \times 0.01 \, \text{moles of CO} = 0.05 \, \text{moles of CO} \]

### Step 4: Calculate the Mass of CO
To find the mass of CO, we need its molecular mass. The molecular mass of CO (carbon monoxide) is:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol
Thus, the molecular mass of CO = 12 + 16 = 28 g/mol.

Now, we can calculate the mass of CO:
\[ \text{Mass of CO} = \text{Number of moles} \times \text{Molecular mass} = 0.05 \, \text{moles} \times 28 \, \text{g/mol} = 1.4 \, \text{g} \]

### Step 5: Calculate the Mass Percentage of CO
The mass percentage of CO in the original mixture is calculated using the formula:
\[ \text{Mass \% of CO} = \left( \frac{\text{Mass of CO}}{\text{Total mass of the mixture}} \right) \times 100 \]

Substituting the values:
\[ \text{Mass \% of CO} = \left( \frac{1.4 \, \text{g}}{2 \, \text{g}} \right) \times 100 = 70\% \]

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

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