# How many moles is 135 L of ammonia gas at STP?

Aug 19, 2016

$6.02 \text{moles}$

#### Explanation:

Since we're at STP, we can use the ideal gas law equation.
$P \times V = n \times R \times T$.

• P represents pressure (could have units of atm, depending on the units of the universal gas constant)
• V represents volume (must have units of liters)
• n represents the number of moles
• R is the universal gas constant (has units of $\frac{L \times a t m}{m o l \times K}$)
• T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables:

$\textcolor{red}{\text{Known variables:}}$
$P$
$V$
$R$
$T$

$\textcolor{b l u e}{\text{Unknown variables:}}$
$n$

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 $\frac{L \times a t m}{m o l \times K}$

Now we have to rearrange the equation to solve for n:

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

Plug in the given values:

n = (1cancel"atm"xx135cancelL)/(0.0821(cancel"Lxxatm")/(molxxcancel"K")xx273cancel"K"
$n = 6.02 m o l$