Question #e1374

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" ")|)#