For the reaction: S(g) + O2(g) -----> SO2 (g)
a)How many moles of S and O2 react to form 0.5 mole of SO2? answer: 0.5 moles
b)What mass of SO2 is formed by reacting 64.2 g of S with O2? answer: 96.2 g
c!!) What volume of SO2 forms when 0.963 g of S reacts with O2 at r.t.p? (I need help with this part and the answer will be 0.72 dm^3) show your working 
thaaaaaaanks in advanceeeeee
a) if the no. of moles SO2 formed when S and O2 react=0.5 moles
then since the mole ratio is 1:1 the no.of moles of S and O2 are also 0.5 mole
b) i think ur answer for this is wrong, because u took the wrong Ar's
mass=no.of moles*Ar
no.of moles of S=64.2/32=2.00625 moles
mass of S02 produced=no.of moles*Mr
Mr of SO2=32+(16*2)=64
therefore the mass=64*2.00625
mass=128.4g
c)no.of moles=vol at rtp/24 dm^3
no.of moles of sulphur=mass/Ar
no.of moles=0.963/32
no. of moles=0.03
therefore vol at rtp=24*0.03=0.72 dm^3
