NaBr- used to produce AgBr for use in photography can itself be prepared as follows-Fe - Br 2- F eBr 2F eBr 2- Br 2- Fe 3 Br 8 not balancedF e 3 Br 8- N a 2 CO 3- N aBr - CO 2- F e 3 O 4 not balancedHow much F e- in kg- is consumed to produce 4-12 - 103 kg NaBr

On converting the given amount of NaBr into mole = WM = 4.12 × 10^3 × 10^3g(23 + 80) = 4.12 × 10^6103 mol Balanced reaction nbsp; (i) (Fe + Br_2 → FeBr_2) × 3 ...

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Standard XII

Chemistry

Stoichiometric Calculations

NaBr, used to...

Question

$N a B r$ , 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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Solution

On converting the given amount of $N a B r$ into mole $= \frac{W}{M} = \frac{4.12 \times 10^{3} \times 10^{3} g}{(23 + 80)} = \frac{4.12 \times 10^{6}}{103} m o l$
Balanced reaction
$(i) (F e + B r_{2} \to F e B r_{2}) \times 3$
$\Rightarrow 3 F e + 3 B r_{2} \to 3 F e B r_{2}$
$(i i) 3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8}$
$(i i i) F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4}$
From stoichiometry of all the reaction 8 mole of $N a B r$ is obtained from 1 mole of $F e_{3} B r_{8} .$ Which is obtained from 3 mole of $F e B r_{2}$ in previous reaction.
According to first reaction 3 mole of $F e B r_{2}$ is obtained from 3 mole of $F e$ . Hence, we can say that 8 mole of $N a B r$ will be obtained from 3 mole of $F e$ .
$∵$ 8 mole of $N a B r$ is obtained from = 3 mole of $F e$
$∴ \frac{4.12 \times 10^{6}}{103}$ mole of $N a B r$ will be obtained from $= \frac{3}{8} \times \frac{4.12 \times 10^{6}}{103}$ mole of $F e$
$∴$ wt. of required $F e = n \times m o l a r m a s s$
$\frac{3}{8} \times \frac{4.12 \times 10^{6}}{103} \times 56 = 840 k g$

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Similar questions

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

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}

Q. Conc.

H_{2} S O_{4}

can not be used to prepare pure

H B r

from

N a B r

as it reacts slowly with

N a B r

Q. Mass of iron required to produce

2.06 \times 10^{3}

N a B r

is :

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NaBr, used to produce AgBr for use in photography can itself be prepared as follows: Fe+Br2→FeBr2 FeBr2+Br2→Fe3Br8 (not balanced) Fe3Br8+Na2CO3→NaBr+CO2+Fe3O4 (not balanced) How much Fe, in kg, is consumed to produce 4.12×103 kg NaBr