Question

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

Answer 1

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`