A) 0.5 L
B) 1 L
C) 2.5 L
D) 0.1 L
Correct Answer: C
Solution :
Key Idea: While calculating, keep in mind about limiting agent. The limiting agent is the reagent which is present in lesser quantity and is consumed during the reaction. The quantity of limiting reagent decides the amount of product. \[\underset{4\,\,L}{\mathop{N{{H}_{3}}}}\,+\underset{1.5\,\,L}{\mathop{HCl}}\,\to \underset{solid}{\mathop{N{{H}_{4}}Cl}}\,\] According to reaction, 4 L of \[N{{H}_{3}}\] will react with 4 L of \[HCl\] \[\because \] Only 1.5 L \[HCl\] is given. \[\therefore \] \[HCl\] is the limiting reagent. \[\therefore \] 1.5 L of \[N{{H}_{3}}\] will react with 1.5 L of \[HCl\]to form \[N{{H}_{4}}Cl\]. \[\therefore \] Volume of gaseous mixture after reaction i.e., amount of \[N{{H}_{3}}\] left unreacted \[=4.0-1.5=2.5\,L\]
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