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56.0 \mathrm{~L} of nitrogen gas is mixed with excess of hydrogen gas and it is found that \mathrm{20 \mathrm{~L}} of ammonia gas is produced. The volume of unused nitrogen gas is found to be ___ \mathrm{L}.

Option: 1

46


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\\\mathrm{N}_2+\mathrm{H}_2 \longrightarrow 2 \mathrm{NH}_3 \\\; \; \; \; \; \; \; \; \; \; \;

          \mathrm{(excess)}

2 mol of \mathrm{NH}_3 formed from 1 mol of \mathrm{N}_2

2 \times 22.4 \; \mathrm{L \; of\; NH_{3}}= 22.4 \text{ L of N}_2 \text{ required}

20 \; \mathrm{L \; of\; NH_{3}}= 10 \text{ L of N}_2 \text{ required}

So, from 56.0 L of \mathrm{NH_{3}} is used , so 46 L is unused.

Answer = 46

Posted by

Sanket Gandhi

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