# Question e1374

Aug 10, 2017

$475$ ${\text{m}}^{3}$ ${\text{CH}}_{4}$

#### Explanation:

We're asked to find the volume (in ${\text{m}}^{3}$) of ${\text{CH}}_{4}$ required to completely react with $1.9 \times {10}^{6}$ ${\text{L Cl}}_{2}$ at S.T.P.

To do this, know that one mole of an (ideal) gas occupies a volume of $22.4$ $\text{L}$ (if S.T.P. is taken to be $1$ $\text{atm}$ and $273.15$ $\text{K}$).

With this in mind, we can convert the given volume of ${\text{Cl}}_{2}$ to moles:

1.9xx10^6cancel("L Cl"_2)((1color(white)(l)"mol Cl"_2)/(22.4cancel("L Cl"_2))) = color(red)(ul(8.48xx10^4color(white)(l)"mol Cl"_2

Now, we can use the coefficients of the given chemical equation to find the relative number of moles of ${\text{CH}}_{4}$ that react:

color(red)(8.48xx10^4)cancel(color(red)("mol Cl"_2))((1color(white)(l)"mol CH"_4)/(4cancel("mol Cl"_2))) = color(green)(ul(2.12xx10^4color(white)(l)"mol CH"_4

Now, we use the same molar volume principle above to find the number of liters of ${\text{CH}}_{4}$:

color(green)(2.12xx10^4)cancel(color(green)("mol CH"_4))((22.4color(white)(l)"L CH"_4)/(1cancel("mol CH"_4))) = color(purple)(ul(4.75xx10^5color(white)(l)"L CH"_4

Lastly, we convert from $\text{L}$ to "m"^, knowing that

ul(1color(white)(l)"mL" = 1color(white)(l)"cm"^3#:

$\textcolor{p u r p \le}{4.75 \times {10}^{5}} \cancel{\textcolor{p u r p \le}{\text{L"))((10^3cancel("mL"))/(1cancel("L")))((1cancel("cm"^3))/(1cancel("mL")))((1color(white)(l)"m"^3)/(100^3cancel("cm"^3))) = color(blue)(ulbar(|stackrel(" ")(" "475color(white)(l)"m"^3color(white)(l)"CH"_4" }} |}$