To find the mass percentage of \( CO_2 \) in the original mixture of \( CO \) and \( CO_2 \), we will follow these steps:
### Step 1: Write the balanced chemical equation
The reaction between carbon monoxide (\( CO \)) and iodine pentoxide (\( I_2O_5 \)) can be represented as:
\[
5 CO + I_2O_5 \rightarrow 5 CO_2 + I_2
\]
### Step 2: Calculate the moles of \( I_2 \) produced
Given that 2.54 grams of \( I_2 \) were produced, we can calculate the number of moles of \( I_2 \) using its molar mass. The molar mass of \( I_2 \) is 254 g/mol.
\[
\text{Moles of } I_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{2.54 \, \text{g}}{254 \, \text{g/mol}} = 0.01 \, \text{mol}
\]
### Step 3: Determine the moles of \( CO \) consumed
From the balanced equation, we see that 1 mole of \( I_2 \) requires 5 moles of \( CO \). Therefore, the moles of \( CO \) that reacted can be calculated as:
\[
\text{Moles of } CO = 5 \times \text{moles of } I_2 = 5 \times 0.01 = 0.05 \, \text{mol}
\]
### Step 4: Calculate the mass of \( CO \)
The molar mass of \( CO \) is 28 g/mol. Thus, the mass of \( CO \) can be calculated as:
\[
\text{Mass of } CO = \text{moles} \times \text{molar mass} = 0.05 \, \text{mol} \times 28 \, \text{g/mol} = 1.4 \, \text{g}
\]
### Step 5: Calculate the mass of \( CO_2 \)
Since the total mass of the mixture is 2 g, we can find the mass of \( CO_2 \) by subtracting the mass of \( CO \) from the total mass:
\[
\text{Mass of } CO_2 = \text{Total mass} - \text{Mass of } CO = 2 \, \text{g} - 1.4 \, \text{g} = 0.6 \, \text{g}
\]
### Step 6: Calculate the mass percentage of \( CO_2 \)
The mass percentage of \( CO_2 \) in the original mixture can be calculated using the formula:
\[
\text{Mass \% of } CO_2 = \left( \frac{\text{mass of } CO_2}{\text{total mass}} \right) \times 100 = \left( \frac{0.6 \, \text{g}}{2 \, \text{g}} \right) \times 100 = 30\%
\]
Thus, the mass percentage of \( CO_2 \) in the original mixture is **30%**.
