Given:
N2, volume of 1L of NH3 at STP
Required:
Grams of N2
Solution:
N2 + 3H2 -> 2NH3
Molar mass of N2 = 28g
From the ideal gas equation PV = nRT
n = PV/RT
n = (1 atm)(100 L)/(0.08206 L-atm/mol-K)(273K)
n = 4.46 mol of NH3
from the reaction, we need 2 moles of NH3 to get 1 mole of N2
4.46 mol NH3(1 mol N2/2 mol NH3) = 2.23 moles N2
2.23 moles of N2(28 g N2/1 mol N2) = 62.5g N2
