# Question c6b11

Mar 19, 2016

${\text{2.5 kg H}}_{2}$

#### Explanation:

The idea here is that you can use the ideal gas law equation to determine how many moles of hydrogen gas you have in the balloon.

Once you know that, use hydrogen's molar mass to determine how many grams would contain that many moles, then finally convert the grams to kilograms by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 kg" = 10^3"g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

An interesting thing to notice here is that the problem mentions STP - you don't actually need to worry about that, since the number of moles of gas is assumed to be constant.

In other words, the number of moles of gas you used to fill up the balloon at STP is equal to the number of moles of gas present at $\text{658 mmHg}$ and $- {8}^{\circ} \text{C}$.

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

Notice that you must convert the pressure of the gas from mmHg to atm, so use the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 atm " = " 760 mmHg}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Likewise, do not forget that you must absolute temperature, i.e. the temperature in Kelvin.

Rearrange the ideal gas law equation to solve for $n$, the number of moles of gas you have in the balloon

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

Plug in your values to get

$n = \left(\frac{658}{760} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 31000color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * [273.15 + (-8)]color(red)(cancel(color(black)("K}}}}\right)$

$n = {\text{1232.9 moles H}}_{2}$

Use hydrogen gas' molar mass to find the mass of hydrogen that would contain this many moles

1232.9color(red)(cancel(color(black)("moles H"_2))) * "2.01588 g"/(1color(red)(cancel(color(black)("mole H"_2)))) = "2485.4 g"

Expressed in kilograms, the answer will be

"mass of hydrogen gas" = color(green)(|bar(ul(color(white)(a/a)"2.5 kg"color(white)(a/a)|)))#

I will leave the answer rounded to two sig figs.