# What volume of the gas absorbed by the alkali ?

• terainfizik
Now, what happens when CO2 reacts with KOH? And how much CO2 do we have in total?In summary, when a mixture of 10cm^3 of methane and 10cm^3 of ethane is sparked with excess oxygen and cooled to room temperature, the resulting gas is passed through aqueous potassium hydroxide. The CO2 produced from the combustion of methane and ethane will react with the KOH, resulting in the absorption of all or most of the CO2. The balanced equations show that for every 1 mole of methane and 2 moles of ethane combusted, 30 cm^3 of CO2 is produced. Therefore, the total volume of gas absorbed by the alkali is

## Homework Statement

A mixture of 10cm^3 of methane and 10cm^3 of ethane was sparked with an excess oxygen . After cooling to room temperature , the residual gas was passed through aqueous potassium hydroxide

What volume of the gas absorbed by the alkali ?

N/A

## The Attempt at a Solution

I have no idea for the solution . Any hint ?

what are the results of 'sparking' methane and ethane with oxygen? i.e. what are the products of combustion?

Water and carbon dioxide . I think . So..?

And what kind of gas is carbon dioxide? What happens when you pass such a gas through an alkaline solution like KOH?

Gokul43201 said:
And what kind of gas is carbon dioxide? What happens when you pass such a gas through an alkaline solution like KOH?

I have no idea . Guide pls .

KOH is a base. What do bases (or basic/alkaline solutions) react with?

I only know that acid react with base will produce water and salt .

Good, then keep in mind that CO2 is an acidic oxide. It is something like an acid, minus the water. It too, will react with a base, forming a salt. The point of this is to recognize that when you pass a gaseous mixture containing CO2 through a solution of KOH, all (or a very large part of) the CO2 will be removed by this reaction.

Write down and balance the equations for the combustion of ethane and methane. How much CO2 is produced from the given quantities?

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I think the equation is as below :

For methane combustion :
CH4 + O2 ---> CO2 + H2O
Since 1 mole of CH4 : 1 mole of CO2
SO: 10CM^3 CH4 will gives 10 cm^3 CO2

For ethane combustion :
C2H6 + O2 ---> CO2 + H2O
Since 1 mole of C2H6 : 1 mole of CO2
So: 10cm^3 C2H6 will gives 20 cm^3 CO2

Total gas has absorbed is 30 cm^3 ?
Am I correct ?

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You have not balanced the equations. You need to do that before you determine the mole ratios.

terainfizik said:
I think the equation is as below :

For methane combustion :
CH4 + O2 ---> CO2 + H2O
Since 1 mole of CH4 : 1 mole of CO2
SO: 10CM^3 CH4 will gives 10 cm^3 CO2

For ethane combustion :
C2H6 + O2 ---> CO2 + H2O
Since 1 mole of C2H6 : 1 mole of CO2
So: 10cm^3 C2H6 will gives 20 cm^3 CO2

Total gas has absorbed is 30 cm^3 ?
Am I correct ?

{EDIT}
For methane combustion :
CH4 + 3O2 ---> CO2 + 2H2O
Since 1 mole of CH4 : 1 mole of CO2
SO: 10CM^3 CH4 will gives 10 cm^3 CO2

For ethane combustion :
C2H6 + 7/2O2 ---> 2CO2 + 3H2O
Since 1 mole of C2H6 : 2 mole of CO2
So: 10cm^3 C2H6 will gives 20 cm^3 CO2 ?

so
total volume is 30cm^3 ?

Looks good.