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

1 Answer
Jun 20, 2016

Answer:

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.
#PxxV = nxxRxxT#.

  • 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 #(Lxxatm)/ (molxxK)#)
  • 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(g)#. 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 #(Lxxatm)/ (molxxK)#

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

# n = (PV)/(RT)#

#n = (1cancel"atm"xx15cancelL)/(0.0821(cancel"Lxxatm")/(molxxcancel"K")xx273cancel"K"#
#n = 0.6692mol#

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#