# Question 22b90

Aug 12, 2017

(a) $0.167$ $\text{mol H"_2"S}$

(b) $3.73$ $\text{L H"_2"S}$

#### Explanation:

• (a) the number of moles of $\text{H"_2"S}$ that react

• (b) the volume of $\text{H"_2"S}$ (at S.T.P.) that would react

(a)

We're given that $8.00$ ${\text{g O}}_{2}$ react, so let's use the molar mass of ${\text{O}}_{2}$ ($31.999$ $\text{g/mol}$) 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 $\text{H"_2"S}$ that react:

$\textcolor{red}{0.250} \cancel{\textcolor{red}{\text{mol O"_2))((2color(white)(l)"mol H"_2"S")/(3cancel("mol O"_2))) = color(blue)(ulbar(|stackrel(" ")(" "0.167color(white)(l)"mol H"_2"S"" }} |}$

$\text{ }$

(b)

At standard temperature and pressure (if it is taken as $1$ $\text{atm}$ and $273.15$ $\text{K}$), 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 $\text{H"_2"S}$ to liters:

$\textcolor{b l u e}{0.167} \cancel{\textcolor{b l u e}{\text{mol H"_2"S"))((22.4color(white)(l)"L H"_2"S")/(1cancel("mol H"_2"S"))) = color(darkblue)(ulbar(|stackrel(" ")(" "3.73color(white)(l)"L H"_2"S"" }} |}$