If you have 20L of nitrogen, and 25L of hydrogen, how many molecules of NH3 can you make? How much excess reactant is left over?

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