Correct Answer - B
`2SO_(2)+O_(2)rarr2SO_(3)`
`Rate=-(d[O_(2)])/(dt)=(1)/(2)(d[SO_(3)])/(dt)`
The rate of formation of S)_(3) is
`(d[SO_(3)])/(dt)=100g min^(-1)=(100)/(80) mol min^(-1)`
`[because "molar mass" of SO_(3)=80 g " mol"^(-1)]`
`(d[O_(2)])/(dt)=(1)/(2)((100)/(80))mol min^(-1)`
`(d[O_(2)])/(dt)=(1)/(2)((100)/(80))xx32gmin^(-1)`
`[because "molar mass of "O_(2)=32g" mol"^(-1)]`
`=20g" min"^(-1)`