How many grams of n2 are required to produce 100.0 l 0f nh3 at stp?

The ideal gas equation is PV = nRT. n = PV/RT n =
(1 atm)(100 L)/(0.08206 L-atm/mol-K)(273K)
 n = 4.46 mol of NH3,we need 2 moles of NH3 to get 1 mole of N2 4.46 mol NH3 so
(1 mol N2/2 mol NH3) = 2.23 moles N2 2.23 moles of N2(28 g N2/1 mol N2) =

Answer is 62.5g N2

What’s the question

The answer is letter c