Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
52 views
in Chemistry by (96.8k points)
closed by
When `22.4 L` of `H_(2)(g)` is mixed with 11.2 of `Cl_(2)(g)`, each at STP, the moles of HCl(g) formed is equal to
A. 1 mole of HCL(g)
B. 2 moles of HCl (g)
C. `0.5` mole of HCl (g)
D. `1.5` moles of HCl (g)

1 Answer

0 votes
by (96.8k points)
selected by
 
Best answer
Correct Answer - A
The limiting reacgent gives the moles og product formed
in the reaction.
`underset(22.4L)(H_(2)(g)) + underset(11.2 L)(Cl_(2) (g)) rarr underset(2" mol")(2HCl(g))`
`because 22.4` volume at STP is occupied by
`therefore 11. 2` L volume will be occupied by
`Cl_(2) = (1xx11.2)/22.4 "mol"=0.5` mol
Thus, `underset(1" mol")(H_(2)(g))+underset(0.5" mol")(Cl_(2)(g)) rarr 2HCl(g)`
Since, `Cl_(2)` possesses minimum number of moles, thus
it is the limiting reagent.
As per equation , mol `Cl_(2)=2" mol HCl"`
`therefore 0.5 ` mol `Cl_(2) = 2 xx 0.5` mol HCl
= `1.0` mol HCl
Hence,`1.0` mole of HCl (g) is produced by `0.5` mole of
`Cl_(2) ` (or `11.2`L).

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...