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

A

420 g

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B

4200 g

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C

251 g

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D

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

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

N a_{2} C O_{3}

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

)

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}

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. If the yield of reaction (ii) is

60 %

and reaction (iii) is

70 %

, then mass of iron required to produce

2.06 \times 10^{3}

N a B r

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}

N a_{2} C O_{3}

are in excess?
(molar masses of

N a, B r

F e

are respectively

23, 80

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}

C u S O_{4} .5 H_{2} O

having 40 % purity by moles undergoes the following sequence of reactions in a flask having large amounts of

K C N

C u S O_{4} .5 H_{2} O \to C u S O_{4} + 5 H_{2} O (Yield = 100%)

C u S O_{4} + 2 K C N \to C u (C N)_{2} + K_{2} S O_{4} (Yield = 80%)

2 C u (C N)_{2} \to 2 C u C N + (C N)_{2} (Yield = 60%)

2 C u C N + 6 K C N \to 2 K_{3} [C u (C N)_{4}] (Yield = 50%)

Find the moles of impure

C u S O_{4} .5 H_{2} O

required for producing 28.45 g of the complex.
(Molar mass of Cu = 63.5 g/mol; molar mass of K = 39 g/mol)

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