# What is the volume of 75.0 g of O_2 at STP?

Aug 9, 2017

$V = 105 L$

#### Explanation:

$p = 1 a t m$
$t = 273.15 K$
$R = 0.08206 \frac{L \cdot a t m}{m o l \cdot K}$
n=(75.0gO_2)/(16.00(g/(mol)O_2)=4.6875mol

Ideal gas law: $P V = n R T$, solve for volume:
$V = \frac{n R T}{P}$

$V = \frac{4.6875 m o l \cdot 0.08206 \frac{L \cdot a t m}{m o l \cdot K} \cdot 273.15 K}{1 a t m}$

$V = 105 L$