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Fluorides of Xenon

For the react...

Question

For the reaction, $2 F e (N O_{3})_{3} + 3 N a_{2} C O_{3} \to F e_{2} (C O_{3})_{3} + 6 N a N O_{3}$
Initially if $2.5$ mole of $F e (N O_{3})_{2}$ and $3.6$ mole of $N a_{2} C O_{3}$ is taken. If $6.3$ mole of $N a N O_{3}$ is obtained then $%$ yield of given reaction is:

50 %

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84 %

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87.5 %

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100 %

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Solution

The correct option is C $87.5 %$
Given reaction :
$2 F e {(N O_{3})}_{3} + 3 N a_{2} C O_{3} \to F e_{2} {(C O_{3})}_{3} + 6 N a N O_{3}$
Therefore,
Two moles of $F e (N O_{3})_{3}$ reacts with three moles of $N a_{2} C O_{3}$
Therefore,
$2.5$ moles of $F e (N O_{3})_{3}$ reacts with $\frac{3}{2} \times 1.5 = 3.75$ moles of $N a_{2} C O_{3}$
But $3.6$ moles moles of $N a_{2} C O_{3}$ are given.
(It means that the same amount of $F e (N O_{3})_{3}$ will be nonreactive)
From the equation,
$3$ moles of $N a_{2} C O_{3}$ gives $6$ moles of $N a N O_{3}$
Therefore,
$3.6$ moles of $N a_{2} C O_{3}$ will give $7.2$ moles of $N a N O_{3}$
Hence,
% yield $= \frac{e x p e r i m e n t a l y i e l d}{c a l c u l a t e d y i e l d}$
$= \frac{6.3}{7.2} \times 100$

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