Chem q

The vanadium in 50.0 cm³ of a 0.800 mol dm- solution of NH.VO, reacts with 506
cm of sulfur(lV) oxide gas measured at 20.0 °C and 98.0 kPa
Use this information to calculate the oxidation state of the vanadium in the solution
after the reduction reaction with sulfur(lV) oxide
Explain your working.
The gas constant R=8.31 J K-' mol-

I don't undertand why they did 0.020366 times 2. I get up to 0.020366 then I don't get the rest

Because when sulphur dioxide is used as a reducing agent, it is oxidised to sulphate ions (they needed you to just know this for this question - in an actual exam, they'd give you the half equation or tell you outright).
Essentially, the oxidation state changes from +4 to +6, which involves each SO2 molecule losing 2 electrons, hence why it is doubled to find the moles of electrons released