Question #b61a4
1 Answer
Explanation:
Start by using the ideal gas law equation to find the total number of moles of gas present in the mixture.
#color(blue)(ul(color(black)(PV = nRT)))#
Here
#P# is the pressure of the gas#V# is the volume it occupies#n# is the number of moles of gas present in the sample#R# is the universal gas constant, equal to#0.0821("atm L")/("mol K")# #T# is the absolute temperature of the gas
Rearrange to solve for
#PV = nRT implies n = (PV)/(RT)#
Plug in your values to find
#n = (2.46 color(red)(cancel(color(black)("atm"))) * 5.0color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (27 + 273.15)color(red)(cancel(color(black)("K"))))#
#n ~~ "0.5 moles"#
Now, you know that this mixture is
#28.4 color(red)(cancel(color(black)("g mixture"))) * overbrace("84.5 g C"/(100color(red)(cancel(color(black)("g mixture")))))^(color(blue)("= 84.5% carbon")) = "23.998 g C"#
Use the molar mass of carbon to determine how many moles of carbon are present in the sample
#23.998 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 1.998 ~~ "2 moles C"#
Notice that the two gases contain equal numbers of moles of carbon, since
#"C"_x"H"_8 implies "1 mole C"_x"H"_8 = xcolor(white)(.)"moles C"# #"C"_x"H"_10 implies "1 mole C"_x"H"_10 = xcolor(white)(.)"moles C"#
Now, if you take
#(0.5 - n)color(white)(.)"moles C"_x"H"_10#
You can also say that the sample contains
#overbrace((n * x)color(white)(.)"moles C")^(color(blue)("from C"_x"H"_8)) + overbrace([(0.5 - n) * x ]color(white)(.)"moles C")^(color(blue)("from C"_x"H"_10)) = "2 moles C"#
This is equivalent to
#color(red)(cancel(color(black)(n * x))) + 0.5 * x - color(red)(cancel(color(black)(n * x))) = 2#
which gets you
#0.5x = 2 implies x= 2/0.5 = 4#
Therefore, you can say that the two gases are
#"C"_4"H"_8" "# and#" " "C"_4"H"_10#