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

B

Explanation

`underset("Initial 5 mol")(A_(2)) + underset(" 7 mol")(3B_(2)) rarr underset(0)(2 AB_(3))`
By reactionn 1 mol `A_(2)` requires 3 mol `B_(2)`
Hence, 7 mol `B_(2)` will reacts with `7/3` moles `A_(2)`
`A_(2) ` left`= 5 - 7/3 = 8//3 ` mol `A_(2)` (g)
1 mol `A_(2)` produces 2 mol `AB_(3)`
Hence, `7/3` mol ` A_(2)` will produce `7/3xx 2` mol `AB_(3)`
Total mol of gases in vessel
`= 8/3 mmol `A_(2) (g) + 14/3` mol `AB_(3)` (g)
`= 22/3` mol