# #15cm^3# of a mixture of ethane and methane were completely oxidized by #43cm^3# of oxygen. The volumes were measured at the same temperature and pressure. What was the composition of the mixture?

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

The idea here is that when dealing gases that are being kept under the *same conditions* for pressure and temperature, you can say that the **mole ratios** that exists between them are **equivalent** to **volume ratios**.

Start by writing the balanced chemical equations that describe these combustion reactions.

You will have

#"CH"_ (4(g)) + color(blue)(2)"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O"_ ((l))#

for the combustion of methane and

#"C"_ 2"H"_ (6(g)) + color(purple)(7/2)"O"_ (2(g)) -> 2"CO"_ (2(g)) + 3"H"_ 2"O"_ ((l))#

for the combustion of ethane.

Now, notice that **every mole** of methane that takes part in the reaction consumes **moles** of oxygen gas. Similarly, **every mole** of ethane that takes part in the reaction consumes **moles** of oxygen gas.

This is equivalent to saying that

#"1 cm"^3color(white)(.)"methane consumes"color(white)(.)color(blue)(2)color(white)(.)"cm"^3# #"O"_2#

#"1 cm"^3color(white)(.)"ethane consumes"color(white)(.)color(purple)(7/2)color(white)(.)"cm"^3# #"O"_2#

If you take

#xcolor(red)(cancel(color(black)("cm"^3))) + ycolor(red)(cancel(color(black)("cm"^3))) = 15 color(red)(cancel(color(black)("cm"^3)))#

which gets you

#x + y = 15#

and that

#(color(blue)(2) * x) color(red)(cancel(color(black)("cm"^3))) + (color(purple)(7/2) * y) color(red)(cancel(color(black)("cm"^3))) = 43color(red)(cancel(color(black)("cm"^3)))#

which gets you

#2x + 7/2y = 43#

Use the first equation to write

#x = 15 - y#

Plug this into the second equation to get

#2 * (15 - y) + 7/2y = 43#

Solve for

#4 * (15 - y) + 7y = 86#

#60 - 4y + 7y = 86#

#3y = 26 implies y = 26/3#

Consequently, you will have

#x = 15 - 26/3 = 19/3#

You can thus say that the initial mixture contained

#19/3 = 6.3color(white)(.)"cm"^3 -> "CH"_4#

#26/3 = 8.7color(white)(.)"cm"^3 ->"C"_2"H"_6#

The answers are rounded to two **sig figs**, the number of sig figs you have for your values.