N2(g) +3H2(g) →2NH3(g)
Given Initially , amount of N2 = 2 moles
amount of H2 = 2 moles
When equillibrium is reached,
Amount of N2 used = 0.5 moles
Thus according to the balanced chemical reaction
1 mole of N2 on reaction will produce = 2 moles of NH3
Thus 0.5 mole of N2 on reaction will produce = 2 ×0.5 moles of NH3 = 1 mole of NH3
Thus at equillibrium 1 mol of NH3 will be present.