Question

NaBr, used to produce AgBr for use in photography can be self prepared as follows:

$Fe + Br_2 \longrightarrow FeBr_2$ ....(i)

$FeBr_2 + Br_2 \longrightarrow Fe_3Br_8$ ....(ii) (not balanced)

$Fe_3Br_8 + Na_2CO_3 \longrightarrow NaBr + CO_2 + Fe_3O_4$ ....(iii) (not balanced)

(At. mass : Fe = 56, Br = 80)

16. Mass of iron required to produce 2.06 × $10^3$ kg NaBr

• A. 420 gm
• B. 420 kg
• C. 4.2 × $10^5$ kg
• D. 4.2 × $10^8$ gm

Answer 1

Answer: (B)

420 kg

Answer 2

The problem involves a series of chemical reactions to produce NaBr. We need to balance the reactions and use stoichiometry, considering reaction yields where specified.

1. Balancing the Chemical Equations:

• Reaction (i): $Fe + Br_2 \longrightarrow FeBr_2$ (Already balanced)
• Reaction (ii): $FeBr_2 + Br_2 \longrightarrow Fe_3Br_8$ To balance Fe, multiply $FeBr_2$ by 3: $3FeBr_2 + Br_2 \longrightarrow Fe_3Br_8$. Now count Br: $3 \times 2 + 2 = 8$ on the left, 8 on the right. Balanced. Balanced (ii): $3FeBr_2 + Br_2 \longrightarrow Fe_3Br_8$
• Reaction (iii): $Fe_3Br_8 + Na_2CO_3 \longrightarrow NaBr + CO_2 + Fe_3O_4$ Balance Br: 8 Br on left, so 8 NaBr on right: $Fe_3Br_8 + Na_2CO_3 \longrightarrow 8NaBr + CO_2 + Fe_3O_4$. Balance Na: 8 Na on right, so 4 $Na_2CO_3$ on left: $Fe_3Br_8 + 4Na_2CO_3 \longrightarrow 8NaBr + CO_2 + Fe_3O_4$. Balance C: 4 C on left, so 4 $CO_2$ on right: $Fe_3Br_8 + 4Na_2CO_3 \longrightarrow 8NaBr + 4CO_2 + Fe_3O_4$. Check O: $4 \times 3 = 12$ on left. $4 \times 2 + 4 = 12$ on right. Balanced. Balanced (iii): $Fe_3Br_8 + 4Na_2CO_3 \longrightarrow 8NaBr + 4CO_2 + Fe_3O_4$

2. Overall Stoichiometry of Fe to NaBr: From (i): $1 \text{ mol Fe} \longrightarrow 1 \text{ mol } FeBr_2$

From (ii): $3 \text{ mol } FeBr_2 \longrightarrow 1 \text{ mol } Fe_3Br_8$

From (iii): $1 \text{ mol } Fe_3Br_8 \longrightarrow 8 \text{ mol } NaBr$

Combining these: To get 1 mol $Fe_3Br_8$, we need 3 mol $FeBr_2$. To get 3 mol $FeBr_2$, we need 3 mol Fe. So, $3 \text{ mol Fe} \longrightarrow 1 \text{ mol } Fe_3Br_8$. Since 1 mol $Fe_3Br_8$ gives 8 mol NaBr, then: $3 \text{ mol Fe} \longrightarrow 8 \text{ mol NaBr}$

3. Molar Masses: Fe = 56 g/mol Br = 80 g/mol Na = 23 g/mol NaBr = 23 + 80 = 103 g/mol

4. Solution for Q.16: Mass of iron required to produce 2.06 × $10^3$ kg NaBr (100% yield)

1. Convert target NaBr mass to grams: Mass of NaBr = 2.06 × $10^3$ kg = 2.06 × $10^3 \times 10^3$ g = 2.06 × $10^6$ g.
2. Calculate moles of NaBr: Moles of NaBr = Mass / Molar mass = (2.06 × $10^6$ g) / (103 g/mol) = 2 × $10^4$ mol.
3. Calculate moles of Fe required (from overall stoichiometry): From $3 \text{ mol Fe} \longrightarrow 8 \text{ mol NaBr}$: Moles of Fe = (3/8) × Moles of NaBr = (3/8) × (2 × $10^4$ mol) = (3/4) × $10^4$ mol = 0.75 × $10^4$ mol = 7500 mol.
4. Calculate mass of Fe required: Mass of Fe = Moles of Fe × Molar mass of Fe = 7500 mol × 56 g/mol = 420000 g.
5. Convert mass of Fe to kg: Mass of Fe = 420000 g / 1000 g/kg = 420 kg.

The final answer is $\boxed{\text{420 kg}}$.