N2(g)+3H2(g) ->2NH3(g) how many liters of H2 are required to produce 47 miles of NH3

Answer 1

Assuming the reaction is run at STP (standard temperature and pressure)....

47 molles NH₃ x 3 moles H₂/2 moles NH₃ = 70.5 moles H₂ required

At STP, 1 mole of an ideal gas = 22.4 L, thus...

70.5 moles H₂ x 22.4 L/mole = 1579 liters = 1600 liters (2 sig. figs.)