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Limiting Reagent or Reactant

If yield of t...

Question

If yield of third reaction is $90$ percent, then moles of $C O_{2}$ formed, when $2.06 \times 10^{3}$ gm $N a B r$ is formed, is:

A

20

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B

10

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C

40

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D

none

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Solution

The correct option is B $10$
Balanced chemical equations are -
$F e + B r_{2} \to F e B r_{2}$
$3 F e B r_{2} + B r_{2} \to F e_{3} B r_{8}$
$F e_{3} B r_{8} + 4 N a_{2} C O_{3} \to 8 N a B r + 4 C O_{2} + F e_{3} O_{4}$
When $8$ moles of $N a B r$ are formed, $4$ moles of carbon dioxide are formed.
The number of moles of carbon dioxide formed is $\frac{2.06 \times 10^{3}}{102.9} \times \frac{4}{8} = 10$ .

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Similar questions

Q. If yield of reaction (iii) is

90 %

, then kilomoles of

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N a B r

Q. If the yield of second reaction is

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C a (H C O_{3})_{2}

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