In the preparation of iron from haematite `(Fe_(2)O_(3))` by the reduction with carbon
`Fe_(2)O_(3)+CtoFe+CO_(2)`
how much 80% pure iron may be produced from 120 kg of 90% pure `Fe_(2)O_(3)`?

Options

A. 94.5 kg
B. 60.48 kg
C. 116.66 kg
D. 120 kg

Answer

To solve the problem of how much 80% pure iron can be produced from 120 kg of 90% pure hematite (Fe₂O₃), we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction for the reduction of hematite with carbon is:
\[ \text{Fe}_2\text{O}_3 + 3\text{C} \rightarrow 2\text{Fe} + 3\text{CO}_2 \]

### Step 2: Calculate the effective mass of Fe₂O₃
Given that we have 120 kg of 90% pure Fe₂O₃, we first need to find the mass of pure Fe₂O₃:
\[ \text{Mass of pure Fe}_2\text{O}_3 = 120 \, \text{kg} \times \frac{90}{100} = 108 \, \text{kg} \]

### Step 3: Convert the mass of Fe₂O₃ to grams
To work with moles, we convert kilograms to grams:
\[ 108 \, \text{kg} = 108,000 \, \text{g} \]

### Step 4: Calculate the number of moles of Fe₂O₃
The molar mass of Fe₂O₃ is calculated as follows:
- Fe: 56 g/mol (2 atoms)
- O: 16 g/mol (3 atoms)
\[ \text{Molar mass of Fe}_2\text{O}_3 = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \, \text{g/mol} \]

Now, we can calculate the number of moles of Fe₂O₃:
\[ \text{Moles of Fe}_2\text{O}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{108,000 \, \text{g}}{160 \, \text{g/mol}} = 675 \, \text{moles} \]

### Step 5: Use stoichiometry to find moles of Fe produced
From the balanced equation, we see that 2 moles of Fe₂O₃ produce 4 moles of Fe. Therefore, the ratio is:
\[ \frac{4 \, \text{moles Fe}}{2 \, \text{moles Fe}_2\text{O}_3} = 2 \]

Thus, the moles of Fe produced from 675 moles of Fe₂O₃ is:
\[ \text{Moles of Fe} = 675 \, \text{moles Fe}_2\text{O}_3 \times \frac{4 \, \text{moles Fe}}{2 \, \text{moles Fe}_2\text{O}_3} = 1350 \, \text{moles Fe} \]

### Step 6: Convert moles of Fe to mass
The molar mass of Fe is 56 g/mol, so the mass of Fe produced is:
\[ \text{Mass of Fe} = 1350 \, \text{moles} \times 56 \, \text{g/mol} = 75,600 \, \text{g} \]

### Step 7: Calculate the mass of 80% pure iron
Since the iron produced is 80% pure, we need to find the effective mass of pure iron:
\[ \text{Mass of pure Fe} = 75,600 \, \text{g} \times \frac{80}{100} = 60,480 \, \text{g} \]

### Step 8: Convert grams to kilograms
Finally, we convert the mass of pure iron to kilograms:
\[ 60,480 \, \text{g} = 60.48 \, \text{kg} \]

### Final Answer
The amount of 80% pure iron produced from 120 kg of 90% pure Fe₂O₃ is **60.48 kg**.

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