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Introduction to Le Chatelier's Principle

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 2.5 mole of $F e (N O_{3})_{2}$ and 3.6 mole of $N a_{2} C O_{3}$ are taken. If 6.3 mole of $N a N O_{3}$ is obtained then % yield of given reaction is:

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87.5

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100

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Solution

The correct option is B 87.5
$2 F e {(N O_{3})}_{3} + 3 {N a}_{2} {C O}_{3} ⟶ {F e}_{2} {(C O_{3})}_{3} + 6 N a {N O}_{3} i n i t i a l 2.5 3.6 - - a f t e r 2.5 - 2.4 = 0.1 0 1.2 6.3 r e a c t i o n$
As 2 moles of $F e (N O_{3})_{3}$ reacts with 3 moles of $N a_{2} C O_{3}$ .
Thus 2.5 moles of $F e (N O_{3})_{3}$ reacts with= $\frac{3}{2} \times 2.5$ moles of $N a_{2} C O_{3}$ =3.75 moles of $N a_{2} C O_{3}$
As $N a_{2} C O_{3}$ present is 3.6 moles only. Thus, $N a_{2} C O_{3}$ is limiting reagent.
Now, 3 moles of $N a_{2} C O_{3}$ gives=6 moles of $N a N O_{3}$
3.6 moles of $N a_{2} C O_{3}$ gives= $\frac{6}{3} \times 3.6$ moles of $N a N O_{3}$
3.6 moles of $N a_{2} C O_{3}$ gives=7.2 moles of $N a N O_{3}$
But $N a N O_{3}$ obtained is 6.3 moles.
Thus % yield= $\frac{6.3}{7.2} \times 100 = 87.5 %$

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