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

the answer is true.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be