For the reaction `Fe_(2)O_(3)+3CO to 2Fe+3CO_(2)`, the volume of carbon monoxide required to reduce one mole of ferric oxide is

Options

A. `67.2 dm^(3)`
B. `11.2 dm^(3)`
C. `22.4 dm^(3)`
D. `44.8 dm^(3)`

Explanation

To solve the problem, we need to determine the volume of carbon monoxide (CO) required to reduce one mole of ferric oxide (Fe₂O₃) according to the given reaction:

**Reaction:**
\[ \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \]

### Step-by-Step Solution:

1. **Identify the Stoichiometry of the Reaction:**
From the balanced chemical equation, we see that 1 mole of Fe₂O₃ reacts with 3 moles of CO. This means that to reduce 1 mole of ferric oxide, we need 3 moles of carbon monoxide.

2. **Determine the Volume of CO at STP:**
At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies a volume of 22.4 liters. Therefore, to find the volume occupied by 3 moles of CO, we can use the following calculation:
\[ \text{Volume of CO} = \text{Number of moles of CO} \times \text{Volume occupied by 1 mole of gas} \]
\[ \text{Volume of CO} = 3 \text{ moles} \times 22.4 \text{ L/mole} \]

3. **Calculate the Total Volume:**
\[ \text{Volume of CO} = 3 \times 22.4 = 67.2 \text{ liters} \]

4. **Convert to Decameter Cubes (if needed):**
Since 1 decameter cube (dm³) is equivalent to 1 liter, we can express the volume as:
\[ 67.2 \text{ liters} = 67.2 \text{ dm}^3 \]

5. **Final Answer:**
Therefore, the volume of carbon monoxide required to reduce one mole of ferric oxide is **67.2 liters** or **67.2 dm³**.

### Summary:
The volume of carbon monoxide required to reduce one mole of ferric oxide (Fe₂O₃) is **67.2 liters**.

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