For reaction `A(g)` + `B(g)` we start with `2` moles of `A` and `B` each. At equilibrium `0.8` moles of `AB` is formed. Then how much of `A` changes to `AB` in `%` ------ will be

Options

A. `20%`
B. `40%`
C. `60%`
D. `4%`

Answer

B

Explanation

`{:(,,A(g),+,B(g),hArr,AB(g)),(,"Initial mole",2,,2,,0),(,"At equilibrium",2-x,,2-x,,x=0.8):}`
`:. %` of A changed to AB = `(0.8)/(2)xx100=40%`