# Question e26c0

May 15, 2016

Here's what I got.

#### Explanation:

The idea here is that you need to use the ideal gas law equation to find the total number of moles present in the container, then use the molar masses of the three species to determine how many grams of each you have.

So, the ideal gas law equation looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Rearrange it to solve for $n$, the total number of moles present in the container

$P V = n R T \implies n = \frac{P V}{R T}$

Plug in your values to find -- do not forget to convert the temperature from degrees Celsius to Kelvin

n = (5 color(red)(cancel(color(black)("atm"))) * 10 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (25 + 273.15)color(red)(cancel(color(black)("K")))) = "2.043 moles gas"#

Now, you know that you equal numbers of moles of each gas, which means you can find the number of moles of oxygen gas, ${\text{O}}_{2}$, carbon dioxide, ${\text{CO}}_{2}$, and hydrogen gas, ${\text{H}}_{2}$, by dividing the total number of moles by $3$

${n}_{{O}_{2}} = {n}_{C {O}_{2}} = {n}_{{H}_{2}} = \text{2.043 moles"/3 = "0.681 moles}$

Now use the molar mass of each chemical species to convert the number of moles to grams

$0.681 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)(|bar(ul(color(white)(a/a)"30. g CO}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$0.681 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)(|bar(ul(color(white)(a/a)"22 g O}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$0.681 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles H"_2))) * "2.016 g"/(1color(red)(cancel(color(black)("mole H"_2)))) = color(green)(|bar(ul(color(white)(a/a)"1.4 g H}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answers rounded to two sig figs.