How many grams of N2 are required to produce 100.0 L of NH3 at STP? 6.25 g

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