If 96 g so2 is added to 2 moles of oxygen at Stp, calculate the volume of so3 that is formed
mols SO2 added =g/molar mass = 96/64 = 1.5
mold SO2 = 2 mols
2SO2 + O2 ==> 2SO3
..1.5.......2.0...........?
First, this is a limiting reagent (LR) problem. Determine the identity of the LR.
How many moles O2 are necessary to react with all of the SO2.
That's 1.5 mols SO2 x (1.0 mols O2/ 2 mols O2) = 1.5 x 1/2 = 0.75 moles SO2. Do you have that much O2. Yes, you have more than enough; therefore, SO2 is the LR.
mols SO3 produced = 1.5 mols SO2 x (2 mols SO3/2 moles SO3) = 1.5 x 2/2 = 1.5 mols SO3