A 20.0-L nickel container was charged with 0.852 atm of xenon gas and 1.34 atm of fluorine gas at 400°C. The xenon and fluorine react to form xenon tetrafluoride. What mass (g) of xenon tetrafluoride can be produced assuming 100% yield?

The chemical reaction for the given scenario can be written as:
Xe + 2F_2 -> XeF_4
That is, 1 mole of xenon reacts with 2 moles of fluorine to produce 1 mole of xenon tetrafluoride.
In this question, 0.852 atm of xenon gas reacts with 1.34 atm of fluorine gas. Using the ideal gas law, we can calculate the moles of each of the reactants.
Therefore, moles of xenon = PV/RT = (0.852 atm x 20 l)/(0.08205 l atm/mole k  x 673 k) 
= 0.309 moles 
(Do remember to convert the temperature to the Kelvin scale.)
Similarly, moles of fluorine = 0.485 moles
Now, using the stoichiometry, we can see that fluorine is in limited quantity since 0.309 moles of xenon will react with 0.618 moles (= 2 x 0.309 moles) of fluorine gas. And hence, the fluorine gas will dictate how much product will be formed.
From the chemical equation, 2 moles of fluorine produces 1 mole of xenon tetrafluoride. And hence the moles of xenon tetrafluoride produced are:
moles of xenon tetrafluoride = 1/2 x 0.485 moles = 0.243 moles
The molecular mass of xenon tetrafluoride is 207.28 gm/moles and hence the total amount of xenon tetrafluoride that is formed is 50.37 gm (= 0.243 moles x 207.28 g/mole).
Hope this helps.