Book Online Demo

Check Your IQ

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

40 ml of a hydrocarbon undergoes combustion in 260ml of oxygen and gives 160ml of CO_2.If all volumes are measured under similar conditions of temperature and pressure, the formula of the hydrocarbon is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

C₄H₈

C₆H₁₄

C₃H₈

C₄H₁₀

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

\large {C_x}{H_y} + \left( {x + \frac{y}{4}} \right){O_2} \to xC{O_2} + \frac{y}{2}{H_2}O\

40\;ml\;\;\;\;\;\large 40\left( {x + \frac{y}{4}} \right)\;\;\;\;\;\;\;\;\;\;\;\40x\

Given

40x = 160

x = 4

\large 40\left( {x + \frac{y}{4}} \right)\ = 260

4x + y = 26

4(4) + y = 26

y = 10

Hydrocarbon is C₄H₁₀

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!