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Stoichiometric Calculations

How many gram...

Question

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 $55.85 {g mol}^{- 1})$
$3 F e + 3 B r_{2} \to 3 F e B r_{2}$
$3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8}$
$F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4}$

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Solution

$3 F e + 3 B r_{2} \to 3 F e B r_{2} (i)$ $3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8} (i i)$
$F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4} (i i i)$

Mol of Fe = 1 mol (given)
In reaction (i), 3 mol of Fe produce 3 mol of $F e B r_{2}$ .
hence, 1 mol will produce 1 mol of $F e B r_{2}$ .

In (ii) reaction, 3 mol of $F e B r_{2}$ produces 1 mol of $F e_{3} B r_{8}$ .
So, 1 mol will produce 0.33 mol.

In (iii) reaction, 1 mol of $F e_{3} B r_{8}$ produces 8 mol of $N a B r$ .
So, 0.33 mol produce = 8 $\times$ 0.33 = 2.64 mol of $N a B r$ .
Amount of $N a B r$ in grams = 2.64 mol $\times$ 103 g/mol = 271.92 g

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Similar questions

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}

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. Number of atoms in

558.5

[Atomic mass of

F e = 55.85 u

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

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