Question

10 mL of a gaseous hydrocarbon was burnt completely in 80 mL of O2 at NTP. The remaining gas
occupied 70 mL at NTP. This volume became 50 mL on treatment with KOH solution. What is the
formula of the hydrocarbon.​

Answer 1

Answer:

C2H4, Ethane

Explanation:

CxHy +(x+y/4)O2→xCO2 +y/2 +y/2H2O

1ml (x+y/4) ml xml

10ml 10(x+y/4)ml 10x ml

Volume of Carbon dioxide =(70−50)=20 ml

10x=20 therefore x=2

Volume of carbon dioxide + volume of oxygen(left) =70 ml

volume of oxygen (left) =70−20=50 ml

Volume of O2 (used)  =80−50=30ml

10(x+y/4)=30

Solve for y, putting x=2,y=4

the formula is C2H4,

The required formula is of ethane.

Answer 2

Answer:

etrerte

Explanation:

According to standard assumptions, in combustion reactions we take water in liquid form. Only $CO_{2}$ is absorbed in KOH.

In the remaining gaseous mixture, there is unused oxygen and $CO_{2}$ which amounts to 70 ml. On treatment with KOH, the $CO_{2}$ will get absorbed. Therefore, total amount of $CO_{2}$ produced is as follows.

          (70 - 50) ml = 20 ml

This means 50 ml of oxygen will be left unused Hence, oxygen used is:

          (80 - 50) ml = 30 ml

At constant temperature and pressure volume is directly proportional to moles. Therefore, we have 10 moles of hydrocarbon which requires 30 moles of oxygen and produces 20 moles of $CO_{2}$.

Also, we have 1 mole of hydrocarbon requires 3 moles of oxygen and produce 2 moles of $CO_{2}$.

Hence, the compound should contain 2 atoms of carbon.

It can be either ethane, ethene or ethyne

Now balancing the reaction we get the compound to be $C_{2}H_{4}$, that is, ethene.

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