Question

Can someone help me with this chemistry question?

40cm^3 of a mixture of methane and ethane undergo combustion with an excess of oxygen gass. after reaction, the product gases were passed through aqueous potassium hydroxide resulting in a contraction in gas volume of 70cm^3. calculate the percent by volume of methane in the mixture. all gas volumes measured in rtp

Answer 1

The mixture contains 25 % methane by volume.

Explanation:

Let `x =` the volume of methane and
Let `y =` the volume of ethane.

Then `bb((1))color(white)(m) x + y = 40`

The aqueous `"KOH"` reacted with the carbon dioxide produced by combustion of the mixture.

`color(white)(l)"CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O"`
`"1 cm"^3color(white)(mmmmll)"1 cm"^3`

From methane:

`"Volume of CO"_2 = x color(red)(cancel(color(black)("cm"^3color(white)(l) "CH"_4))) × ("1 cm"^3color(white)(l) "CO"_2)/(1color(red)(cancel(color(black)("cm"^3color(white)(l) "CH"_4)))) = xcolor(white)(l) "cm"^3color(white)(l) "CO"_2`

`"2C"_2"H"_6 + "7O"_2 → "4CO"_2 + "6H"_2"O"`
`color(white)(l)"2 cm"^3color(white)(mmmmml)"4 cm"^3`

From ethane:

`"Volume of CO"_2 = y color(red)(cancel(color(black)("cm"^3color(white)(l) "C"_2"H"_6))) × ("4 cm"^3color(white)(l) "CO"_2)/(2color(red)(cancel(color(black)("cm"^3color(white)(l) "C"_2"H"_6)))) = 2ycolor(white)(l) "cm"^3color(white)(l) "CO"_2`

Thus,

`bb((2))color(white)(m)x + 2y = 70`

From Equation 1,

`bb((3))color(white)(m) y = 40 - x`

Substitute Equation 3 into Equation 2.

`x + 2(40 - x) = 70`

`x + 80 - 2x = 70`

`x = 10`

The volume of methane is `"10 cm"^3`.

`"Percent methane" = "Volume of methane"/"Total volume" × 100 % = (10 color(red)(cancel(color(black)("cm"^3))))/(40 color(red)(cancel(color(black)("cm"^"3")))) × 100 % = "25 %"`