Verified Solution • 2025–26 Session moderate

QJEEM22Ch5b51: JEE Main 2022 Chemistry - Class 11 JEEM22Ch5b51 Solution: 56.0 L of nitrogen gas is mixed with excess hydrogen gas and it …

56.0 L of nitrogen gas is mixed with excess hydrogen gas and it is found that 20 L of ammonia gas is produced. The volume of unused nitrogen gas is found to be _________ L.

Step-by-Step Verified Solution

N2( g)+3H2( g)→2NH3( g)\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})

Since H2\mathrm{H}_{2} is in excess and 20 L20 \mathrm{~L} of ammonia gas is produced.

Hence, 2 moles NH3≡1\mathrm{NH}_{3} \equiv 1 mole N2(v∝n)\mathrm{N}_{2} \quad(v \propto \mathrm{n})

20 LNH3≡10 L N220 \mathrm{~L} \mathrm{NH}_{3} \equiv 10 \mathrm{~L} \mathrm{~N}_{2}

Volume of N2\mathrm{N}_{2} left =56−10=56-10

=46 L=46 \mathrm{~L}

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