20 seconds after the start of the reaction 2SО2 + О2 → 2SО3 3,2 g remain unoxidised of 4,8 g of sulfur oxide (IV). Find the rate of oxidation of sulfur dioxide .

given:
m1(SО2) = 4,8 g
m2(SО2) = 3,2 g
Δt = 20 sec
find:
υ - ?
Solution:
υ=ΔС/Δt; ΔС= Δν
Δm(SО₂)=4,8 g-3,2 g=1,6 g
М(SО2) = 64 g/mol
ν=m/М=(1,6 g)/(64 g/mol)=0,025 mol
if V = 1 lit, т.
υ=(0,025)/(20 •1)=0,00125mol/lit•s
Answer:   υ =0,00125 mol/lit • sec.