Question

Question #3c541

Answer 1

Here's what I got.

Explanation:

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"`.