On passing carbon dioxide over red hot carbon, carbon monoxide is produced as pe the equation `CO_(2) + C to 2CO`. Calculate the volume of carbon monoxide at STP wher 3 g of carbon is consumed in the reaction.

Explanation

To solve the problem of calculating the volume of carbon monoxide produced at STP when 3 g of carbon is consumed in the reaction \( CO_2 + C \rightarrow 2CO \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The balanced reaction is:
\[ CO_2 + C \rightarrow 2CO \]
This indicates that 1 mole of carbon reacts with 1 mole of carbon dioxide to produce 2 moles of carbon monoxide.

2. **Calculate the number of moles of carbon**:
To find the number of moles of carbon, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of carbon (C) is approximately 12 g/mol. Therefore, for 3 g of carbon:
\[ \text{Number of moles of carbon} = \frac{3 \text{ g}}{12 \text{ g/mol}} = 0.25 \text{ moles} \]

3. **Determine the moles of carbon monoxide produced**:
According to the balanced equation, 1 mole of carbon produces 2 moles of carbon monoxide. Thus, if we have 0.25 moles of carbon:
\[ \text{Moles of CO produced} = 0.25 \text{ moles of C} \times 2 = 0.50 \text{ moles of CO} \]

4. **Calculate the volume of carbon monoxide at STP**:
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of 0.50 moles of carbon monoxide is:
\[ \text{Volume of CO} = 0.50 \text{ moles} \times 22.4 \text{ L/mole} = 11.2 \text{ liters} \]

5. **Final answer**:
The volume of carbon monoxide produced is **11.2 liters**.