#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
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
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.
