Gas law with molecules at the same time How many grams of oxygen are present at 25.0 C in a 5.00L tank of oxygen enriched air? The total pressure is 30.0 atm and the only other gas is nitrogen with a pressure of 17.0 atm.

After converting the temperature to Kalvin, I do not really know what to do. 
How can I find the grams of oxygen with another molecule in the container?
We have 2 gases in the container, oxygen ($O_2$) and nitrogen ($N_2$). Now we are given that $N_2$ has a pressure of $17$ atm out of $30$ atm, the total volume of the container is $5.00$ L and that the temperature is $25^o$C or $298$ K. So somehow, we need to find out the volume that oxygen occupies and from there we will be able to find out the grams of oxygen present in the container.

First let's find out the mole fraction $N_2$.
You can find mole fraction by the following formula,
 $$Mole\, Fraction = \frac{Partial\,Pressure\,of\,Gas}{Total\,Pressure}$$ 

So,
Mole fraction of $N_2$= $\frac{17}{30}$=$0.6$  

Therefore mole fraction of $O_2$=$1-0.6$=$0.4$  

Now we need to find out the total no. of moles by the formula,
$$PV={nRT}$$  

Putting the values given for the whole system,
$$30\times5=n\times0.0821\times298$$
$$\Rightarrow\,n=6.1$$

So moles of oxygen in container = $0.4\times6.1$ = $2.5$
Consequently, grams of $O_2$ present in container = $2.5\times32$ = $78.4$  

So that is how you get the grams of oxygen. Hope you understood everything. If not, feel free to ask by commenting. Good Luck!