A) 0.5 litre
B) 1 litre
C) 2.5 litre
D) 0.1 litre
Correct Answer: C
Solution :
\[\underset{4\,litre}{\mathop{N{{H}_{3}}}}\,+\underset{1.5\,litre}{\mathop{HCl}}\,\xrightarrow{{}}N{{H}_{4}}Cl(s)\] Here, \[HCl\] is the limiting reagent. Hence, 1 of \[N{{H}_{3}}\]will react with 1.5 1 of \[HCl\] \[(1:1)\] to give \[N{{H}_{4}}Cl\] (solid). Hence volume of gaseous mixture after reaction will be = 4 -\[1.5=2.5\]litre (of \[N{{H}_{3}}\])
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