41 g iron and 70 g carbon monoxide are allowed to react to produce iron carbonyl `Fe(CO)_(5).` Find the excess reactant and how much mass of it will be left at the end of reaction :-

Options

A. `CO, 42 gm`
B. `Fe, 42 gm`
C. `Co, 13 gm`
D. `Fe, 13 gm`

Answer

To solve the problem of determining the excess reactant and the amount left after the reaction between iron and carbon monoxide to produce iron carbonyl \( \text{Fe(CO)}_5 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ \text{Fe} + 5 \text{CO} \rightarrow \text{Fe(CO)}_5 \]

### Step 2: Calculate the number of moles of each reactant
1. **Calculate moles of iron (Fe)**:
- Molar mass of iron (Fe) = 56 g/mol
- Given mass of iron = 41 g
\[ \text{Moles of Fe} = \frac{\text{mass of Fe}}{\text{molar mass of Fe}} = \frac{41 \text{ g}}{56 \text{ g/mol}} \approx 0.732 \text{ moles} \]

2. **Calculate moles of carbon monoxide (CO)**:
- Molar mass of carbon monoxide (CO) = 28 g/mol
- Given mass of carbon monoxide = 70 g
\[ \text{Moles of CO} = \frac{\text{mass of CO}}{\text{molar mass of CO}} = \frac{70 \text{ g}}{28 \text{ g/mol}} \approx 2.5 \text{ moles} \]

### Step 3: Determine the limiting reactant
From the balanced equation, we see that 1 mole of Fe reacts with 5 moles of CO.

- For 0.732 moles of Fe, the required moles of CO would be:
\[ 0.732 \text{ moles Fe} \times 5 \text{ moles CO/mole Fe} = 3.66 \text{ moles CO} \]

- We have only 2.5 moles of CO available, which is less than the required 3.66 moles. Therefore, CO is the limiting reactant.

### Step 4: Calculate the amount of excess reactant left
Since CO is the limiting reactant, we will calculate how much iron is left after the reaction.

- Moles of CO consumed:
\[ \text{Moles of CO consumed} = 2.5 \text{ moles} \]

- Moles of Fe required for 2.5 moles of CO:
\[ \text{Moles of Fe required} = \frac{2.5 \text{ moles CO}}{5 \text{ moles CO/mole Fe}} = 0.5 \text{ moles Fe} \]

- Moles of Fe left:
\[ \text{Moles of Fe left} = \text{Initial moles of Fe} - \text{Moles of Fe required} = 0.732 \text{ moles} - 0.5 \text{ moles} = 0.232 \text{ moles} \]

### Step 5: Convert moles of excess Fe back to grams
To find the mass of the excess iron left:
\[ \text{Mass of excess Fe} = \text{Moles of Fe left} \times \text{Molar mass of Fe} = 0.232 \text{ moles} \times 56 \text{ g/mol} \approx 13 \text{ g} \]

### Final Answer
The excess reactant is iron (Fe), and the mass of iron left at the end of the reaction is approximately **13 g**.

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