Magnetite, `Fe_(3)O_(4)`, can be converted into metallic iron by heating with carbon monoxide as represented by this equation:
`Fe_(3)O_(4)(s)+CO(g) rarr Fe(s) +CO_(2)(g)`
The kilograms of `Fe_(3)O_(4)` which must be processed in this way to obtain `5.00 kg` of iron, if the process is `85%` efficient is closest to? `[M: = Fe = 56]`

Options

A. `6.92 kg`
B. `8.15 kg`
C. `20.8 kg`
D. `24.4 kg`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write down the balanced chemical equation
The balanced equation for the conversion of magnetite (Fe₃O₄) to iron (Fe) is:
\[ \text{Fe}_3\text{O}_4(s) + \text{CO}(g) \rightarrow 3\text{Fe}(s) + \text{CO}_2(g) \]

### Step 2: Determine the molar mass of Fe₃O₄
The molar mass of Fe₃O₄ can be calculated as follows:
- Iron (Fe) has a molar mass of 56 g/mol.
- Oxygen (O) has a molar mass of 16 g/mol.

Thus, the molar mass of Fe₃O₄ is:
\[ 3 \times 56 + 4 \times 16 = 168 + 64 = 232 \text{ g/mol} \]

### Step 3: Calculate the moles of iron produced
We are given that we want to produce 5.00 kg of iron. First, convert this mass to grams:
\[ 5.00 \text{ kg} = 5000 \text{ g} \]

Next, calculate the number of moles of iron produced:
\[ \text{Moles of Fe} = \frac{\text{mass}}{\text{molar mass}} = \frac{5000 \text{ g}}{56 \text{ g/mol}} \approx 89.29 \text{ moles} \]

### Step 4: Relate moles of Fe to moles of Fe₃O₄
From the balanced equation, we see that 3 moles of Fe are produced from 1 mole of Fe₃O₄:
\[ 3 \text{ moles of Fe} \rightarrow 1 \text{ mole of Fe}_3\text{O}_4 \]

Thus, the moles of Fe₃O₄ required can be calculated as:
\[ \text{Moles of Fe}_3\text{O}_4 = \frac{\text{Moles of Fe}}{3} = \frac{89.29}{3} \approx 29.76 \text{ moles} \]

### Step 5: Calculate the mass of Fe₃O₄ required
Now, convert moles of Fe₃O₄ to grams:
\[ \text{Mass of Fe}_3\text{O}_4 = \text{Moles} \times \text{Molar mass} = 29.76 \text{ moles} \times 232 \text{ g/mol} \approx 6905.12 \text{ g} \]

Convert this to kilograms:
\[ 6905.12 \text{ g} = 6.905 \text{ kg} \]

### Step 6: Adjust for efficiency
Since the process is 85% efficient, we need to adjust the mass of Fe₃O₄ required:
\[ \text{Actual mass of Fe}_3\text{O}_4 = \frac{\text{Required mass}}{\text{Efficiency}} = \frac{6.905 \text{ kg}}{0.85} \approx 8.12 \text{ kg} \]

### Final Answer
The kilograms of Fe₃O₄ that must be processed to obtain 5.00 kg of iron, considering the 85% efficiency, is approximately **8.12 kg**.

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