Convert 7.9 x 10-5 g to standard notation.

The balanced equation for formation of NH₃ is as follows;
N₂ + 3H₂ --> 2NH₃
Stoichiometry of N₂ to H₂ is 1:3
when gases react, stoichiometry of reactants applies to volumes as well.
Because 1 mol of any substance occupies 22.4 L at STP.
Therefore stoichiometry of volumes of N₂ to H₂ is 1:3
If N₂ is the limiting reactant,
1 L of N₂ reacts with 3 L of H₂
Therefore 20 L of N₂ should react with - 3/1 x 20 = 60 L
Only 25 L of H₂ is present therefore H₂ is the limiting reactant 
ratio of H₂ to NH₃ is 3:2
if 3 L of H₂ forms - 2 L of NH₃
then 25 L of H₂ forms - 2/3 x 25 L = 16.67 L of ammonia 
if 22.4 L of NH₃ contains 1 mol of NH₃
then 16.67 L of NH₃ consists of - 1/22.4 x 16.67 =0.74 mol oh NH₃
1 mol of NH₃ is made of 6.022 x 10²³ NH₃ molecules 
Therefore 0.74 mol of NH₃ made of 6.022 x 10²³ molecules/mol x 0.74 mol 
number of NH₃ molecules formed - 4.45 x 10²³ molecules of NH₃

N₂ is in excess
25 L of H₂ reacts with - 25/3 = 8.33 L 
however 20 L present initially 
Excess reactant left - 20 - 8.33 = 11.67 L of N₂ remaining