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")):}`
How much `Fe` in `kg` is consumed to produce `2.06xx10^(3) kg Nabr " "........(iv)`
If yield of `(iii)` reaction is `90%` then mole of `CO_(2)` formed when `2.06xx10^(3) kg NaBr` is formed.

Options

A. `20`
B. `10`
C. `40`
D. none

Explanation

To solve the problem step by step, we will follow the reactions provided and calculate the required quantities.

### Step 1: Understanding the Reaction
We need to focus on the third reaction given:
\[ \text{Fe}_3\text{Br}_8 + \text{Na}_2\text{CO}_3 \rightarrow \text{NaBr} + \text{CO}_2 + \text{Fe}_3\text{O}_4 \]

### Step 2: Balancing the Reaction
To balance the reaction:
- We have 8 bromine atoms on the left, so we need 8 NaBr on the right.
- This means we need 4 Na2CO3 to balance the sodium.
- We also need to balance the carbon and oxygen.

The balanced equation becomes:
\[ \text{Fe}_3\text{Br}_8 + 4 \text{Na}_2\text{CO}_3 \rightarrow 8 \text{NaBr} + 4 \text{CO}_2 + \text{Fe}_3\text{O}_4 \]

### Step 3: Finding Moles of NaBr
Given that we need to produce \( 2.06 \times 10^3 \) kg of NaBr, we first calculate the number of moles of NaBr produced.

1. **Molar Mass of NaBr**:
- Sodium (Na) = 23 g/mol
- Bromine (Br) = 80 g/mol
- Molar mass of NaBr = 23 + 80 = 103 g/mol

2. **Convert kg to g**:
\[ 2.06 \times 10^3 \text{ kg} = 2.06 \times 10^6 \text{ g} \]

3. **Calculate Moles of NaBr**:
\[ \text{Moles of NaBr} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2.06 \times 10^6 \text{ g}}{103 \text{ g/mol}} \approx 20000 \text{ moles} \]

### Step 4: Finding Moles of CO2 Produced
From the balanced equation, we see that 8 moles of NaBr produce 4 moles of CO2. Therefore, the moles of CO2 produced can be calculated as follows:

1. **Using the ratio from the balanced equation**:
\[ \text{Moles of CO}_2 = \frac{4}{8} \times \text{Moles of NaBr} = \frac{1}{2} \times 20000 = 10000 \text{ moles} \]

### Step 5: Adjusting for Yield
Since the yield of the reaction is 90%, we need to adjust the moles of CO2 produced accordingly.

1. **Calculate the actual moles of CO2 produced**:
\[ \text{Actual moles of CO}_2 = 10000 \times 0.90 = 9000 \text{ moles} \]

### Step 6: Finding the Mass of Fe Consumed
To find the mass of Fe consumed, we need to relate it to the balanced equation.

1. **From the balanced reaction**:
- 1 mole of Fe3Br8 corresponds to 3 moles of Fe.
- Thus, for every 8 moles of NaBr produced, we need 1 mole of Fe3Br8, which requires 3 moles of Fe.

2. **Calculate moles of Fe required**:
\[ \text{Moles of Fe} = \frac{3}{8} \times 20000 = 7500 \text{ moles} \]

3. **Molar Mass of Fe**:
- Iron (Fe) = 56 g/mol

4. **Calculate mass of Fe**:
\[ \text{Mass of Fe} = 7500 \text{ moles} \times 56 \text{ g/mol} = 420000 \text{ g} = 420 \text{ kg} \]

### Final Answers
- Mass of Fe consumed: **420 kg**
- Moles of CO2 formed: **9000 moles**