| ⇦ |
| ⇨ |
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: - Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85g
- If a 0.00001 M solution of HCl is diluted thousand folds the pH of the resulting
- 1,1,2,2-tetrabromoethane on treatment with Zn dust yields an
- Which reagent can be used for the alkylation of aromatic nitro compounds
- The most probable radius (in pm) for finding the electron in He⁺ is
Topics: Some Basic concepts of Chemistry
(5)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85g
- If a 0.00001 M solution of HCl is diluted thousand folds the pH of the resulting
- 1,1,2,2-tetrabromoethane on treatment with Zn dust yields an
- Which reagent can be used for the alkylation of aromatic nitro compounds
- The most probable radius (in pm) for finding the electron in He⁺ is
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