A mixture of 26 ml of H2 and 24ml of O2 is exploded . After cooling the tube contains unreacted

2H2 26ml262=13Limiting reragent + O2 24ml241=24 → 2H2O Volume of H2 required double than O2.26 vol of H2 reat with 13 vol of O2 .Vol of O2 left = 24-13=11 vol = 11 ml .

A mixture of 26 ml of H2 and 24ml of O2 is exploded . After cooling the tube contains unreacted

2H2 26ml262=13Limiting reragent + O2 24ml241=24 → 2H2O Volume of H2 required double than O2.26 vol of H2 reat with 13 vol of O2 .Vol of O2 left = 24-13=11 vol = 11 ml .

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A mixture of 26 ml of H₂ and 24ml of O₂ is exploded . After cooling the tube contains unreacted

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

2 ml of H₂

14 ml of H₂

23 ml of H₂

11 ml of O₂

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\underset{\underset{\frac{26}{2} = 13 Limiting reragent}{26 ml}}{2 H_{2}} + \underset{\underset{\frac{24}{1} = 24}{24 ml}}{O_{2}} \overset{}{\to} 2 H_{2} O$