In a closed vessel, 5 moles of `A_(2)(g)` and 7 moles of `B_(2)` (g) are reacted in the following maner,
`A_(2)(g)+(3B_(2)(g)to2AB_(3)(g)`
What is the total number of moles of gases present in the container at the end of the reaction?

Options

A. `22//3`
B. `7//3`
C. `14//3`
D. `8//3`

Answer

A

Explanation

`underset("Initial 5 mol")(A_(2))+underset(7"mol")(3B_(2))tounderset(0)(2AB_(3))`
By reactiion 1 mole `A_(2)` required 3 moles `B_(2)`.
Hence, 7 moles `B_(2)` will reacts with `7/3` moles `A_(2)`
`A_(2)` left `=5-7/3=8//3` moles `A_(2)(g)`
1 mole `A_(2)` produces 2 moles `AB_(3)`.
Hence, `7/3` moles `A_(2)` will produce
`7/3xx2` moles `AB_(3)`
Total moles of gases in vessel
`=8/3` moles `A_(2)(g)+14/3` moles `AB_(3)(g)`
`=22/3` moles