Find the mass of "2.0 L" of helium gas at STP? The molar mass is "4.002602 g/mol".

Oct 27, 2015

To do this problem, we can use the ideal gas law, knowing that we are at STP:

$P \overline{V} = R T$

where $\overline{V} = \frac{V}{n}$, the molar volume.

For an ideal gas at STP:

• $P = \text{1 bar}$ (standard pressure)
• barV_"ideal" = ?
• $R = \text{0.083145 L" cdot "bar/mol" cdot "K}$
• $T = \text{273.15 K}$ (standard temperature)

We can solve for the ideal molar volume ${\overline{V}}_{\text{ideal}}$, and then from that determine the $\text{mol}$s of helium and thus the mass.

${\overline{V}}_{\text{ideal}} = \frac{R T}{P}$

$= \frac{0.083145 \cdot 273.15}{1} = \textcolor{g r e e n}{\text{22.711 L/mol}}$

Then, we know that when assuming helium is ideal:

${\overline{V}}_{\text{ideal" = barV_"He}}$

"22.711 L"/"1 mol ideal gas" = "2.0 L"/(n_"He")

${n}_{\text{He" = "0.0881 mol}}$

Lastly, converting to grams:

0.0881 cancel"mol He" xx ("4.002602 g")/(cancel"mol He")

$\textcolor{b l u e}{\text{= 0.3525 g He}}$

Oct 27, 2015

The mass of helium present is $\text{0.36 g}$.

Explanation:

$\text{STP}$ is $\text{273.15 K}$ and $\text{1 atm}$.

You will use the ideal gas law in order to determine moles of helium. Then you will multiply the moles of helium times its molar mass to determine the mass of helium in grams.

Ideal gas law

$P V = n R T$, where $P$ is pressure, $V$ is volume, $n$ is moles, $R$ is the gas constant, a $T$ is the temperature.

Known/Given
$P = \text{1 atm}$
$V = \text{2.0 L}$
$R = \text{0.08206 L atm K"^(-1) "mol"^(-1)}$
$T = \text{273.15 K}$
Molar mass of helium: $\text{4.002602 g/mol}$

Unknown
Moles of helium:${n}_{\text{He}}$
Mass of helium in grams

Equations

$P V = n R T$

n_"He"xx(4.002602"g He")/(1"mol He")

Solution

Rearrange the ideal gas law equation to isolate $n$, then solve.

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

n=(1cancel"atm"*2.0cancel"L")/(0.08206cancel"L atm K"^(-1)"mol"^(-1)*273.15cancel"K")="0.089 mol He"

$0.089 \cancel{\text{mol He"xx(4.002602"g He")/(1cancel"mol He")="0.36 g He}}$