What is the mass, in grams, of 15 L of O_2 at STP?

Jun 20, 2016

There are 21 g of ${O}_{2}$.

Explanation:

Since we are at STP this means that we have to 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. Our only unknown is the number of moles of ${O}_{2} \left(g\right)$. Our known variables are P,V,R, and T.

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:

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

n = (1cancel"atm"xx15cancelL)/(0.0821(cancel"Lxxatm")/(molxxcancel"K")xx273cancel"K"
$n = 0.6692 m o l$

To get the mass of ${O}_{2}$, we just have to use the molar mass of oxygen as a conversion factor:

0.6692cancel"mol" O_2 xx (32.00g)/(1cancel "mol") = 21g ${O}_{2}$