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Percentage Composition

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 $3$ moles of $F e (N O_{3})_{3}$ and $3.6$ moles of $N a_{2} C O_{3}$ is taken and $6.3$ moles of $N a N O_{3}$ is obtained, then the $%$ 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 $87.5 %$
$\begin{matrix} 2 F e (N O_{3})_{3} + & 3 N a_{2} C O_{3} & \to F e_{2} (C O_{3})_{3} & + & 6 N a N O_{3} Mole & 3 & 3.6 Mole/Stoichiometric coefficient & 1.5 & 1.2 \end{matrix}$
Limiting reagent is $N a_{2} C O_{3}$ .
So, moles of $N a N O_{3}$ should be formed $= 3.6 \times 2 = 7.2$
$% yield = \frac{6.3}{7.2} \times 100 = 87.5 %$ .

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