Question #22b90

We're asked to find

• (a) the number of moles of #"H"_2"S"# that react

• (b) the volume of #"H"_2"S"# (at S.T.P.) that would react

(a)

We're given that #8.00# #"g O"_2# react, so let's use the molar mass of #"O"_2# (#31.999# #"g/mol"#) to calculate the number of moles of oxygen that react:

#8.00cancel("g O"_2)((1color(white)(l)"mol O"_2)/(31.999cancel("g O"_2))) = color(red)(ul(0.250color(white)(l)"mol O"_2#

Now, let's use the coefficients of the chemical equation to find the relative number of moles of #"H"_2"S"# that react:

#color(red)(0.250)cancel(color(red)("mol O"_2))((2color(white)(l)"mol H"_2"S")/(3cancel("mol O"_2))) = color(blue)(ulbar(|stackrel(" ")(" "0.167color(white)(l)"mol H"_2"S"" ")|)#

#" "#

(b)

At standard temperature and pressure (if it is taken as #1# #"atm"# and #273.15# #"K"#), one mole of an (ideal) gas occupies a volume of #ul(22.4color(white)(l)"L"#.

Using that conversion factor, we can convert from moles of #"H"_2"S"# to liters:

#color(blue)(0.167)cancel(color(blue)("mol H"_2"S"))((22.4color(white)(l)"L H"_2"S")/(1cancel("mol H"_2"S"))) = color(darkblue)(ulbar(|stackrel(" ")(" "3.73color(white)(l)"L H"_2"S"" ")|)#