.........N2 + 3H2 ==> 2NH3
I........4......9.......0
C
E
If we take 4 N2 that will use up 3*4 = 12 molecules of H2 BUT we don't have 12 molecules; therefore, H2 must be the limiting reagent so in th table we will use H2
...........N2 + 3H2 ==> 2NH3
I..........4.....9.........0
C.........-3....-9........6
E..........1.....0........6
Nitrogen and hydrogen combine at high temperature, in the presence of a catalyst, to produce ammonia.
N2(g)+3H2(g)-->2NH3(g)
Assume 4 molecules of nitrogen and 9 molecules of hydrogen are present.
After complete reaction, how many molecules of ammonia are produced?
How many molecules of H2 remain?
How many molecules of N2 remain?
What is the limiting reactant?
2 answers
Molecules of H2: 3 molecules.
Nitrogen atoms present: 3 atoms
Moles of NH3 formed: 2 moles
Nitrogen atoms present: 3 atoms
Moles of NH3 formed: 2 moles