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