Question #3c541

Your starting point here will be the balanced chemical equations for the combustion of these two gases, methane, #"CH"_4#, and ethylene, #"C"_2"H"_4#.

#"C"_2"H"_text(4(g]) + color(red)(3)"O"_text(2(g]) -> 2"CO"_text(2(g]) + 2"H"_2"O"_text((g])#

#"CH"_text(4(g]) + color(blue)(2)"O"_text(2(g]) -> "CO"_text(2(g]) + 2"H"_2"O"_text((g])#

Take a look at the mole ratios that exist between the two gases and oxygen. You will see that

• every mole of ethylene will require #color(blue)(3)# moles of oxygen gas
• every mole of methane will require #color(red)(2)# moles of oxygen gas

Now, let's assume that #x# represents the number of moles of ethylene and #y# the number of moles of methane. You know that

#x + y = "0.3 moles" " " " "color(purple)((1))#

Now, since no mention of pressure and temperature was made, I"ll assume that you're working at STP, Standard Temperature and Pressure.

At STP conditions, which are defined as a pressure of #"100 kPa"# and a temperature of #0^@"C"#, one mole of any ideal gas occupies exactly #"22.7 L"# - this is known as the molar volume of a gas at STP.

Use the molar volume of a gas to find how many moles of oxygen were needed for the combustion of the mixture

#15.68 color(red)(cancel(color(black)("L"))) * "1 mole O"_2/(22.7color(red)(cancel(color(black)("L")))) = "0.69075 moles O"_2#

So, if #x# represents the number of moles of ethylene in the mixture, you know that the combustion of this compound required

#x color(red)(cancel(color(black)("moles C"_2"H"_4))) * (color(red)(3)" moles O"_2)/(1color(red)(cancel(color(black)("mole C"_2"H"_4)))) = color(red)(3)x" moles O"_2#

Likewise, if #y# represents the number of moles of methane, you can say that

#y color(red)(cancel(color(black)("moles C"_2"H"_4))) * (color(blue)(2)" moles O"_2)/(1color(red)(cancel(color(black)("mole C"_2"H"_4)))) = color(blue)(2)y" moles O"_2#

This means that you have

#color(red)(3)x + color(blue)(2)y = "0.69075 moles O"_2" " " "color(purple)((2))#

Since you need to find the mass of methane present in the mixture, solve these two equations for #y#. Use equation #color(purple)((1))# to find

#x = 0.3 - y#

Plug this into equation #color(purple)((2))# to get

#3 * (0.3 - y) + 2y = 0.69075#

#0.9 -3y + 2y = 0.69075#

#y = 0.20925#

So, your initial mixture contained #"0.20925 moles"# of methane. Use methane's molar mass to determine how many grams of methane would contain this many moles

#0.20925 color(red)(cancel(color(black)("moles CH"_4))) * "16.0425 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "3.357 g"#

I'll leave the answer rounded to two sig figs

#m_(CH_4) = color(green)("3.4 g")#

SIDE NOTE Many sources still use the old definition of STP, at which pressure is equal to #"1 atm"# and temperature to #0^@"C"#.

Under these conditions for pressure and temperature, one mole of any ideal gas occupies #"22.4 L"#. If this is the value that you are supposed to use, simply redo the calculations using #"22.4 L"# instead of #"22.7 L"#.