# L have 56 L of H_2 gas at STP. Haw many grams do I have of H_2 gas?

Dec 15, 2015

The mass of $\text{56 L H"_2}$ at $\text{STP}$ is $\text{5.0 g}$.

#### Explanation:

Use the ideal gas law to determine moles of $\text{H"_2}$. Then multiply the moles times the molar mass. $\text{STP=273.15 K and 100 kPa}$.

Part 1: Ideal Gas Law
$P V = n R T$, where $n$ represents moles and $R$ is the gas constant.

Given/Known
$P = \text{100 kPa}$
$V = \text{56 L}$
$R = \text{8.3144598 L kPa K"^(-1) "mol"^(-1)}$
$T = \text{273.15 K}$

Unknown
$n$

Solution
Rearrange the equation to isolate $n$ and solve.

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

n=(100cancel"kPa"xx56cancel"L")/(8.3144598cancel"L" cancel"kPa" cancel"K"^(-1) "mol"^(-1)xx273.15cancel"K")="2.5 mol H"_2" (rounded to two significant figures)

Part 2: Mass of $\text{H"_2}$
Multiply the moles of hydrogen gas times its molar mass, $\text{2.01588 g/mol}$.

$2.5 \cancel{\text{mol H"_2xx(2.01588"g H"_2)/(1cancel"mol H"_2)="5.0 g H"_2}}$