5 moles of `SO_(2)` and 5 moles of `O_(2)` are allowed to react to form `SO_(3)` in a closed vessel. At the equilibrium stage `60%` of `SO_(2)` is used up. The total number of moles of `SO_(2), O_(2)` and `SO_(3)` in the vessel now is

Options

A. `10.0`
B. `8.5`
C. `10.5`
D. `3.9`

Answer

B

Explanation

`SO_(2)(g) +1/2O_(2)(g) hArr SO_3(g)`
`{:(at t = 0,5,5,0),("(moles)",,,),("at equili"^m,5-(5xx0.6),5-(5xx0.3),5xx0.6),("(moles)",2,3.5,3):}`
Total moles = 2 + 3.5 + 3 = 8.5moles .