We can combine the given equations........
$S(s) + 3/2O_2(g) rarr SO_3(g)$
And this clearly gives us the mass transfer; 1 equiv of sulfur is oxidized by 3/2 equiv dioxygen gas.....
$"Moles of sulfur"=(5.00*g)/(32.06*g*mol^-1)=0.156*mol$
And so we need $0.156*molxx3/2xx32.00*g*mol^-1=7.49*g$ with respect to dioxygen gas......
And we use the Ideal Gas Equation.......
$V=(nRT)/P=((7.49*g)/(32.00*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx623.2*K)/(5.25*atm)$
$=??L$