Comprehension # 3
`NaBr`, used to produce `AgBr` for use in photography can be self prepared as follows :
`{:(,Fe+Br_(2)rarrFeBr_(2)," "..........(i),),(,FeBr_(2)+Br_(2)rarrFe_(3)Br_(8)," " ..........(ii) ," "("not balanced")),(,Fe_(3)Br_(8)+Na_(2)CO_(3)rarrNaBr+CO_(2)+Fe_(3)O_(4) , " ".......(iii)," "("not balanced")):}`
If the yield of (ii) is `60% &` (iii) reaction is `70%` then mass of ion required to produce `2.06xx10^(3) kg NaBr`.

Options

A. `10^(5) kg`
B. `10^(5) g`
C. `10^(3) kg`
D. none

Explanation

To solve the problem step by step, we need to follow the reactions provided and account for the yields of the reactions. Here’s how we can approach the solution:

### Step 1: Write down the balanced equations
The reactions provided in the question are:
1. \( \text{Fe} + \text{Br}_2 \rightarrow \text{FeBr}_2 \) (i)
2. \( \text{FeBr}_2 + \text{Br}_2 \rightarrow \text{Fe}_3\text{Br}_8 \) (ii)
3. \( \text{Fe}_3\text{Br}_8 + \text{Na}_2\text{CO}_3 \rightarrow \text{NaBr} + \text{CO}_2 + \text{Fe}_3\text{O}_4 \) (iii)

First, we need to balance the third equation:
- The balanced equation is:
\[ \text{Fe}_3\text{Br}_8 + 4 \text{Na}_2\text{CO}_3 \rightarrow 8 \text{NaBr} + 4 \text{CO}_2 + \text{Fe}_3\text{O}_4 \]

### Step 2: Calculate moles of NaBr
Given that we want to produce \( 2.06 \times 10^3 \) kg of NaBr, we first convert this mass to moles:
- The molar mass of NaBr is \( 23 + 80 = 103 \, \text{g/mol} \).
- Convert kg to g: \( 2.06 \times 10^3 \, \text{kg} = 2.06 \times 10^6 \, \text{g} \).
- Calculate moles of NaBr:
\[ \text{Moles of NaBr} = \frac{2.06 \times 10^6 \, \text{g}}{103 \, \text{g/mol}} \approx 20000 \, \text{mol} \]

### Step 3: Account for yields
From the balanced equation (iii), we see that 1 mole of \( \text{Fe}_3\text{Br}_8 \) produces 8 moles of NaBr. Therefore, to find the moles of \( \text{Fe}_3\text{Br}_8 \) needed:
\[ \text{Moles of } \text{Fe}_3\text{Br}_8 = \frac{20000 \, \text{mol NaBr}}{8} = 2500 \, \text{mol} \]

Considering the yield of the reaction (iii) is 70%, we calculate the actual moles of \( \text{Fe}_3\text{Br}_8 \) needed:
\[ \text{Moles of } \text{Fe}_3\text{Br}_8 \text{ (actual)} = \frac{2500 \, \text{mol}}{0.70} \approx 3571.43 \, \text{mol} \]

### Step 4: Calculate moles of FeBr2
From the balanced equation (ii), 1 mole of \( \text{FeBr}_2 \) produces 1 mole of \( \text{Fe}_3\text{Br}_8 \). Therefore, the moles of \( \text{FeBr}_2 \) needed:
\[ \text{Moles of } \text{FeBr}_2 = 3571.43 \, \text{mol} \]

Considering the yield of the reaction (ii) is 60%, we calculate the actual moles of \( \text{FeBr}_2 \) needed:
\[ \text{Moles of } \text{FeBr}_2 \text{ (actual)} = \frac{3571.43 \, \text{mol}}{0.60} \approx 5952.38 \, \text{mol} \]

### Step 5: Calculate moles of Fe
From the balanced equation (i), 3 moles of Fe produce 1 mole of \( \text{FeBr}_2 \). Therefore, the moles of Fe needed:
\[ \text{Moles of Fe} = 5952.38 \, \text{mol} \times 3 \approx 17857.14 \, \text{mol} \]

### Step 6: Calculate mass of Fe
The molar mass of Fe is approximately \( 56 \, \text{g/mol} \):
\[ \text{Mass of Fe} = 17857.14 \, \text{mol} \times 56 \, \text{g/mol} \approx 1000000 \, \text{g} = 1000 \, \text{kg} \]

### Final Answer
The mass of iron required to produce \( 2.06 \times 10^3 \) kg of NaBr is approximately **1000 kg**.

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