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

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

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)