Question

Question #e1374

Answer 1

`475` `"m"^3` `"CH"_4`

Explanation:

We're asked to find the volume (in `"m"^3`) of `"CH"_4` required to completely react with `1.9xx10^6` `"L Cl"_2` at S.T.P.

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

With this in mind, we can convert the given volume of `"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 `"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 `"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 `"L"` to `"m"^`, knowing that

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

`color(purple)(4.75xx10^5)cancel(color(purple)("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" ")|)`