Question

Question #22b90

Answer 1

(a) `0.167` `"mol H"_2"S"`

(b) `3.73` `"L H"_2"S"`

Explanation:

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"" ")|)`