The balanced equation for the preparation of NaBr is as follows-Fe-Br2- FeBr2 -i yield- 100 3FeBr2-Br2- Fe3Br8-ii yield- 100 Fe3 Br8-4Na2 CO3- 8NaBr-4CO2-Fe3 O4-iii yield- 60 Calculate the mass of iron required to produce 2-06-103 g of NaBr-Considering Na2 CO3 and Br2 to be present in excessmolar mass of Fe-56 g-mol-Br-80 g-mol- Na-23 g-mol

The balanced chemical equations for the preparation of NaBr are as follows: 3Fe+3Br_2→3FeBr_2 nbsp; —(i) 3FeBr_2+Br_2→Fe_3 Br_8 nbsp; — (ii) Fe_3 Br_8+4Na ...

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Byju's Answer

Standard XI

Chemistry

Stoichiometric Calculations

The balanced ...

Question

The balanced equation for the preparation of $N a B r$ is as follows:
$F e + B r_{2} \to F e B r_{2} \dots(i) (yield: 100 %)$
$3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8} ...(ii) (yield: 100 %)$
$F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4} \dots(iii) (yield: 60 %)$
Calculate the mass of iron required to produce $2.06 \times 10^{3} g$ of $N a B r$ .
(Considering $N a_{2} C O_{3}$ and $B r_{2}$ to be present in excess)
(molar mass of $F e = 56 g/mol, B r = 80 g/mol, N a = 23 g/mol$ )

420 g

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4200 g

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251 g

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700 g

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Solution

The correct option is D $700 g$
The balanced chemical equations for the preparation of $N a B r$ are as follows:
$3 F e + 3 B r_{2} \to 3 F e B r_{2} ---(i)$
$3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8} ----(ii)$
$F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4} ---(iii)$
From the balanced chemical equations, it can be seen that 3 mol of $F e$ produces 3 mol of $F e B r_{2}$ which in turn will produce 1 mol of $F e_{3} B r_{8}$ .
$\Rightarrow$ Number of moles of $F e$ required = 3 $\times$ number of mol of $F e_{3} B r_{8}$ required
The molar masses of $F e_{3} B r_{8}$ and $N a B r$ are $808 g/mol$ and $103 g/mol$ respectively.
Since, 8 mol of $N a B r$ is produced from 1 mol of $F e_{3} B r_{8}$
$(8 \times 103) g$ of $N a B r$ is produced from $(1 \times 808)$ of $F e_{3} B r_{8}$
$2.06 \times 10^{3} g N a B r$ will be produced by $= \frac{1 \times 808}{8 \times 103} \times 2.06 \times 10^{3} g$ of $F e_{3} B r_{8}$
Mass of $F e_{3} B r_{8}$ required = $2020 g$
Since, the yield of the third reaction is 60%, then the actual amount of $F e_{3} B r_{8}$ required will be:
Amount of $F e_{3} B r_{8}$ (required) = $\frac{2020}{0.6} = 3366.67 g$
$\Rightarrow$ Amount of $F e$ (required)= $= \frac{3 \times 56 \times 3366.67}{808} = 700 g$
Therefore, the actual amount of Fe required will be equal to $700 g$ .

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Similar questions

Q. The balanced equation for the preparation of

N a B r

is as follows:

F e + B r_{2} \to F e B r_{2} \dots(i) (yield: 100 %)

3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8} ...(ii) (yield: 100 %)

F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4} \dots(iii) (yield: 60 %)

Calculate the mass of iron required to produce

2.06 \times 10^{3} g

N a B r

.
(Considering

N a_{2} C O_{3}

and

B r_{2}

to be present in excess)
(molar mass of

F e = 56 g/mol, B r = 80 g/mol, N a = 23 g/mol

)

Q. The balanced equation for the preparation of

N a B r

is as follows:

F e + B r_{2} \to F e B r_{2} \dots(i) (yield: 100 %)

3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8} ...(ii) (yield: 100 %)

F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4} \dots(iii) (yield: 60 %)

Calculate the mass of iron required to produce

2.06 \times 10^{3} g

N a B r

.
(Considering

N a_{2} C O_{3}

and

B r_{2}

to be present in excess)
(molar mass of

F e = 56 g/mol, B r = 80 g/mol, N a = 23 g/mol

)

Q. How many grams of

N a B r

is produced if 55.85 g of

F e

is consumed in the following series of reactions assuming that

B r_{2}

and

N a_{2} C O_{3}

are in excess?
(molar masses of

N a, B r

and

F e

are respectively

23, 80

and

55.85 {g mol}^{- 1})

3 F e + 3 B r_{2} \to 3 F e B r_{2}

3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8}

F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4}

N a B r

, used to produce

A g B r

for use in photography can itself be prepared as follows:

F e + B r_{2} \to F e B r_{2}

F e B r_{2} + B r_{2} \to F e_{3} B r_{8}

(not balanced)

F e_{3} B r_{8} + N a_{2} C O_{3} \to N a B r + C O_{2} + F e_{3} O_{4}

(not balanced)
How much

F e

, in kg, is consumed to produce

4.12 \times 10^{3} k g N a B r

N a B r

, used to produce

A g B r

for use in photography can itself be prepared as follows:

F e + B r_{2} \to F e B r_{2}

F e B r_{2} + B r_{2} \to F e_{3} B r_{8}

(not balanced)

F e_{3} B r_{8} + N a_{2} C O_{3} \to N a B r + C O_{2} + F e_{3} O_{4}

(not balanced)
How much

F e

, in kg, is consumed to produce

4.12 \times 10^{3} k g N a B r

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