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Structural Isomerism

If 340 g of...

Question

If $340 g$ of mixture of $N_{2}$ and $H_{2}$ in the correct ratio gave a $20 %$ yield of $N H_{3}$ . The produced mass of $N H_{3}$ would be:

A

16 g

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B

17 g

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C

20 g

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D

68 g

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Solution

The correct option is D $68 g$
According to the problem:
Given, the mixture is $340 g m$ ,
Yield % $=$ $20$

$N_{2} + 3 H_{2} \to 2 N H_{3}$

Therefore,
At NTP, $34 g m$ of reaction mixture is equal to the $34 g m$ of $N H_{3}$

Hence, $340 g m$ of reaction mixture is,
$= \frac{34}{34} \times 340 = 340 g m o f N H_{3}$

Percentage yield $= \frac{o b t a i n e d p r o d u c t m a s s}{e x p c t e d p r o d u c t m a s s} \times 100$

$20 = \frac{o b t a i n e d p r o d u c t m a s s}{340} \times 100$

Hence the obtained product mass is $68$ gram of $N H_{3}$

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