Answer
The volume of $O_2$ gas required is $=2.27L$
Work Step by Step
$S(s)+O_2(g)=SO_2(g)$
$2SO_2(g)+O_2(g)=2SO_3$
or
$SO_2(g)+1/2O_2(g)=SO_3$
So to convert $1mol$ $S$ to produce $1mol$ of $SO_3$ we need $1.5mol$ $O_2$
Here we need to completely convert 5.0g $S$ to $SO_3$
Number of $mol$ of $S$ is $n(s)=(5/32.065)mol=0.156mol$
So the required $mol$ of $O_2$ $n(O_2)=(1.5×0.156)mol = 0.234 mol$
Here $P=5.25atm$
$T=(350+273)K= 623K$
So $V=n(O_2)RT/P=(0.234 ×0.082×623)/5.25L=2.27L$
The volume of $O_2$ gas required is $=2.27L$
