28 g of `N_(2)` and 6 g of `H_(2)` were mixed. At equilibrium 17 g `NH_(3)` was produced. The weight of `N_(2)` and `H_(2)` at equilibrium are respectively

Options

A. `11 g`, zero
B. `1g`, `3`
C. `14g`, `3g`
D. `11g`, `3g`

Explanation

`{:(,,N_(2),+,3H_(2),hArr,2NH_(3)),(,,28/28=1,,6/2=3,,0"mole before reaction"),(,,1-1/2,,3-3/2,,17/17=1"mole after reaction"):}`
`:. "Mole of" N_(2)=(1)/(2)`
`Wt. of N_(2)=14 g`
Mole of `H_(2)=(3)/(2)`
Wt. of `H_(2)=(3)/(2)xx2=3g`