Gas law with molecules at the same time
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!