`1` mol of `N_(2)` is mixed with `3` mol of `H_(2)` in a litre container. If `50%` of `N_(2)` is converted into ammonia by the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`, then the total number of moles of gas at the equilibrium are

Options

A. `1.5`
B. `4.5`
C. `3.0`
D. `6.0`

Answer

C

Explanation

`{:(,N_(2)(g),+,3H_(2)(g),hArr,2NH_(3)(g)),("Initial conc.",1,,3,,-),("At. eqm.",0.5,,1.5,,1.0):}`
Total number of moles of equilibrium
`=0.5 +1.5+1.00=3.00`