| ⇦ | 
| ⇨ |
When 22.4 litres of H₂(g) is mixed with 11.2 litres of Cl₂(g) each at S.T.P, the moles of HCl(g) formed is equal to
Options
(a) 1 mol of HCl(g)
(b) 2 mol of HCl(g)
(c) 0.5 mol of HCl(g)
(d) 1.5 mol of HCl(g)
Correct Answer:
1 mol of HCl(g)
Explanation:
H₂ +Cl₂ → 2HCl
t=0 ,22.4,11.2 lit , 0
t=0,or 1 mole, 0.5 mole, 0
at time t (1-0.5) =0.5
0.5×2 = 1 mole
Related Questions: - Which of the following metal has the biggest tendency for the half reaction
- The work done in expansion of an ideal gas from an initial volume v1 to a final v2
- The number of moles of acidified KMnO₄ required to oxidise one mole
- Which expression is correct for the work done in adiabatic reversible expansion
- Which of the following shows electrical conduction?
Topics: Some Basic concepts of Chemistry
(5)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following metal has the biggest tendency for the half reaction
- The work done in expansion of an ideal gas from an initial volume v1 to a final v2
- The number of moles of acidified KMnO₄ required to oxidise one mole
- Which expression is correct for the work done in adiabatic reversible expansion
- Which of the following shows electrical conduction?
Topics: Some Basic concepts of Chemistry (5)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply