Correct Answer - C
`{:(,N_(2),+,3H_(2),hArr,2NH_(1).....(i)),("at t=0",56 g,,8g,,0g),(,=2"mole",,"4 mole",,"0 mole"),("At equilibrium",2-1,,4-3,,34 g),(,=1"mole",,1"mole",,2"moles"):}`
According to eq. (i) 2 moles of ammonia is present and to produce 2 moles of `NH_(3)`, we need 1 mole of `N_(2)` and 3 moles of `H_(2)` hence 2 - 1 = 1 mole of `N_(2)` and 4 - 3 = 1 mole of `H_(2)` are present at equilibrium in vessel.