# How many liters are 3.4 moles of HCl(g) at STP?

##### 1 Answer
Jun 7, 2017

The volume is $\text{77 L}$.

#### Explanation:

This question involves the Ideal Gas Law.

Ideal gas law

$P V = n R T$,

where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvins.

The current STP temperature is ${0}^{\circ} \text{C}$ or $\text{273.15 K}$ (used for gas laws), and the pressure is "10^5 Pa, which most people write as $\text{100 kPa}$ (both pressures are equal).

Organize the data:

Known

$P = \text{100 kPa}$

$n = \text{3.4 mol}$

$R = \text{8.3144598 L kPa K"^(-1) "mol"^(-1)}$
https://en.wikipedia.org/wiki/Gas_constant

$T = \text{273.15 K}$

Unknown: $V$

Solution

Rearrange the equation to isolate $V$. Insert the data and solve.

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

V=(3.4color(red)cancel(color(black)("mol"))xx8.3144598color(white)(.)"L" color(red)cancel(color(black)("kPa")) color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx273.15color(red)cancel(color(black)("K")))/(100color(red)cancel(color(black)("kPa")))="77 L HCl(g)" rounded to two sig figs