What is the percent yield of the following reaction if $42 g$ of $M g C O_{3}$ is heated to give $10 g$ of $M g O$ ?
$M g C O_{3} \to M g O + C O_{2}$

30 %

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40 %

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60 %

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50 %

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Solution

The correct option is D $50 %$
For the given reaction,
$M g C O_{3} \to M g O + C O_{2}$
As per stoichiometry, $1$ mol of $M g C O_{3}$ decomposes to give $1$ mol of $M g O$ and $1$ mol of $C O_{2}$
$I n i t i a l m o l e s o f M g C O_{3} = \frac{42}{84} = 0.5 m o l$
So, theoretically,
$m o l e s o f M g O p r o d u c e d = 0.5 m o l$
$∴ m a s s p r o d u c e d = 0.5 \times m o l a r m a s s o f M g O = 0.5 \times 40 = 20 g$

But the actual yield is $10 g$ of $M g O$ .

So, $P e r c e n t a g e y i e l d = \frac{A c t u a l y i e l d}{T h e o r e t i c a l y i e l d} \times 100$
$= \frac{10 g}{20 g} \times 100 = 50 %$

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