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Question

In the Haber’s process, 30 L of hydrogen and 30 L of nitrogen were taken for the reaction, which yielded only 50% of the expected product. What will be the composition of the gaseous mixture under the aforementioned condition after reaction is completed?

A
20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
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B
10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
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C
20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen
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D
20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
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Solution

The correct option is B 10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
Given: N2+3H22NH3
1 L of nitrogen reacts with 3 L of hydrogen.
30 L of nitrogen will react with 31×30=90 L of hydrogen.
But we are given only 30 L of hydrogen (according to the question)
Therefore, hydrogen is the limiting reagent.
3 L of hydrogen produce 2 L of ammonia.
With 100% efficiency, 30 L of hydrogen will produce 20 L of ammonia
But with 50% efficiency, only 10 L of ammonia is produced.
Therefore, only 15 L of hydrogen reacted.
3 L of hydrogen react with 1 L of nitrogen,
So, 15 L of hydrogen will react with 5 L of nitrogen
Composition of the gaseous mixture in the end will be:
Hydrogen remaining = 30 L - 15 L = 15 L
Nitrogen remaining = 30 L - 5 L = 25 L
Ammonia produced = 10 L

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