To find the mass of iron (Fe) required to produce 2.06 x 10^3 kg of sodium bromide (NaBr), we will follow these steps:
### Step 1: Write the balanced chemical equation
The relevant balanced equation from the provided reactions is:
\[ \text{Fe}_3\text{Br}_8 + \text{Na}_2\text{CO}_3 \rightarrow 8 \text{NaBr} + \text{CO}_2 + \text{Fe}_3\text{O}_4 \]
### Step 2: Determine the molar masses
- Molar mass of NaBr:
\[
\text{Na} = 23 \, \text{g/mol}, \quad \text{Br} = 80 \, \text{g/mol} \quad \Rightarrow \quad \text{Molar mass of NaBr} = 23 + 80 = 103 \, \text{g/mol}
\]
- Molar mass of Fe:
\[
\text{Fe} = 56 \, \text{g/mol}
\]
### Step 3: Calculate the number of moles of NaBr produced
Given that we need to produce 2.06 x 10^3 kg of NaBr:
\[
\text{Mass of NaBr} = 2.06 \times 10^3 \, \text{kg} = 2.06 \times 10^6 \, \text{g}
\]
Now, calculate the moles of NaBr:
\[
\text{Moles of NaBr} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.06 \times 10^6 \, \text{g}}{103 \, \text{g/mol}} \approx 20000 \, \text{mol}
\]
### Step 4: Relate moles of NaBr to moles of Fe
From the balanced equation, we see that:
\[
3 \, \text{moles of Fe} \rightarrow 8 \, \text{moles of NaBr}
\]
Thus, we can set up a ratio to find the moles of Fe needed:
\[
\frac{3 \, \text{moles of Fe}}{8 \, \text{moles of NaBr}} = \frac{x \, \text{moles of Fe}}{20000 \, \text{moles of NaBr}}
\]
Solving for \(x\):
\[
x = \frac{3}{8} \times 20000 \approx 7500 \, \text{moles of Fe}
\]
### Step 5: Calculate the mass of Fe required
Now, we can find the mass of Fe:
\[
\text{Mass of Fe} = \text{Moles of Fe} \times \text{Molar mass of Fe} = 7500 \, \text{mol} \times 56 \, \text{g/mol} = 420000 \, \text{g}
\]
Convert grams to kilograms:
\[
\text{Mass of Fe in kg} = \frac{420000 \, \text{g}}{1000} = 420 \, \text{kg}
\]
### Final Answer
The mass of iron required to produce 2.06 x 10^3 kg of NaBr is **420 kg**.
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