| ⇦ | 
| ⇨ |
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:
1 mole – 0.5 mole = 0.5 mole
2 HCl => 0.5 *2 = 1 mole
Related Questions: - Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for the
- The pair of species that has the same bond order in the following is:
- Treatment of a mixture of CH₃Cl and C₂H₅Cl with sodium in dry ether can generat
- The number of water molecules is maximum in
- The bond angle of sp² – hybrid orbital is
Question Type: Apply
(15)
Difficulty Level: Easy
(1008)
Topics: Basic Concepts of Chemistry
(94)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for the
- The pair of species that has the same bond order in the following is:
- Treatment of a mixture of CH₃Cl and C₂H₅Cl with sodium in dry ether can generat
- The number of water molecules is maximum in
- The bond angle of sp² – hybrid orbital is
Question Type: Apply (15)
Difficulty Level: Easy (1008)
Topics: Basic Concepts of Chemistry (94)
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