A 1.0L container of air (21% O2 & 79%N2) is at STP. Calculate the mass
of nitrogen gas in the container at STP, and the partial pressure of
nitrogen gas at STP.

Hi!!

At constant temperature and volume, the total pressure of a gas sample
is determined by the total number of moles of gas present, whether
this represents a single substance, or a mixture.

The gas law states:
P*V = n*R*T ===> n = (P*V)/(R*T) =
                   = 1 atm * 1 L / (0.0821 L*atm/mol*K * 273.15 K) =
                   = 0.0446 moles of gas (N+O)

Since 79% of the total molecules are Nitrogen molecules we have that there are:
0.79*0.0446 = 0.035234 moles of nitrogen.

The atomic mass of Nitrogen is 14, then the molecular mass is 28,
therefore 1 mol of N2 has a mass of 28 g, then 0.035234 moles of N2
will have a mass of:
0.035234 * 28 g = 0.986552 grams.


At STP the total pressure (P) is 1atm, according to Dalton's Law of
Partial Pressures the partial pressures P_O and P_N will satisfy:
P_O + P_N = P ; 
and the ratio of the partial pressure to the total pressure is equal
to the mole fraction of the component gas, then for this case the
partial pressure of Nitrogen gas is:
P_N = 0.79*P = 0.79*1 atm = 0.79 atm

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For references see "Gas Mixtures and Partial Pressures":
http://wine1.sb.fsu.edu/chm1045/notes/Gases/Mixtures/Gases06.htm


I hope that this helps you. Feel free to request for a clarification
if you need it.

Best regards.
livioflores-ga

thank you! This was very helpful!