Question 3c541

Feb 2, 2016

Here's what I got.

Explanation:

Your starting point here will be the balanced chemical equations for the combustion of these two gases, methane, ${\text{CH}}_{4}$, and ethylene, ${\text{C"_2"H}}_{4}$.

${\text{C"_2"H"_text(4(g]) + color(red)(3)"O"_text(2(g]) -> 2"CO"_text(2(g]) + 2"H"_2"O}}_{\textrm{\left(g\right]}}$

${\text{CH"_text(4(g]) + color(blue)(2)"O"_text(2(g]) -> "CO"_text(2(g]) + 2"H"_2"O}}_{\textrm{\left(g\right]}}$

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 $\textcolor{b l u e}{3}$ moles of oxygen gas
• every mole of methane will require $\textcolor{red}{2}$ 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

$x + y = \text{0.3 moles" " " " } \textcolor{p u r p \le}{\left(1\right)}$

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 $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$, one mole of any ideal gas occupies exactly $\text{22.7 L}$ - 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

$\textcolor{red}{3} x + \textcolor{b l u e}{2} y = \text{0.69075 moles O"_2" " " } \textcolor{p u r p \le}{\left(2\right)}$

Since you need to find the mass of methane present in the mixture, solve these two equations for $y$. Use equation $\textcolor{p u r p \le}{\left(1\right)}$ to find

$x = 0.3 - y$

Plug this into equation $\textcolor{p u r p \le}{\left(2\right)}$ to get

$3 \cdot \left(0.3 - y\right) + 2 y = 0.69075$

$0.9 - 3 y + 2 y = 0.69075$

$y = 0.20925$

So, your initial mixture contained $\text{0.20925 moles}$ 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

${m}_{C {H}_{4}} = \textcolor{g r e e n}{\text{3.4 g}}$

SIDE NOTE Many sources still use the old definition of STP, at which pressure is equal to $\text{1 atm}$ and temperature to ${0}^{\circ} \text{C}$.

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