A container contains 16 g of oxygen ,24g of methane ,96g of ozone and 80 g of so3 and the temp is 227 degree celcius volume is 10L and R is0.08 so calculate the mole fraction and partial pressure of each gas

Dear student

For this question first of all you need to calculate the number of moles of all the components in the container. 
The number of moles are :
oxygen 0.5 mol
methane 0.67 mol
ozone 2 mol
sulphur trioxide 1 mol
​The temperature of the reaction mixture is 500K
R = 0.08
So the total number of moles are: 4.167 mol
Using PV = nRT
The total pressure can be calculated as 
P = 10.668 atm
Now to calculate the mole fraction and the partial pressure use the formula
$$\begin{gathered} mole fraction = \frac{given number of moles}{total number of moles} \\ and the partial pressure = total pressure * mole fraction \\ For ex, for oxygen mole fraction = \frac{0.5}{4.167}= 0.12 \\ and thus partial pressure = 0.12*10.668 = 1.28 atm \end{gathered}$$
The same formula can be used for others also.

Regards