Question

4NH3+5O2 = 4NO + 6 H20

If a container were to have 10 molecules of O2 and 10 molecules of NH3, initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

Answer 1

When both reactants are given you know it is a limiting reagent problem.
10 molecules O2 x (4 moles NH3/5 moles O2)= 8 molecules NH3 needed and you have that much; therefore, oxygen is the limiting reagent but check it to make sure.
10 molecules NH3 x (5 moles O2/4 moles NH3) = 50/4 = 12.5 and you don't have that much oxygen; thus oxygen is the limiting reagent.
8 molecules NH3 will reagent with 10 molecules O2 to form 8 molecules NO. I will leave you with it. Use the coefficients to determine molecules H2O

Answer 2

38 molecules

Answer 3

In the following table we have listed three rows of information. The “Initial” row is the
number of molecules present initially, the “Change” row is the number of molecules that
react to reach completion, and the “Final” row is the number of molecules present at
completion. To determine the limiting reactant, let’s calculate how much of one reactant is
necessary to react with the other.
10 molecules O2 ×
2
3
5 moleculesO
4 moleculesNH
= 8 molecules NH3 to react with all the O2
Because we have 10 molecules of NH3 and only 8 molecules of NH3 are necessary to react
with all the O2, O2 is limiting.
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
Initial 10 molecules 10 molecules 0 0
Change -8 molecules -10 molecules +8 molecules +12 molecules
Final 2 molecules 0 8 molecules 12 molecules
The total number of molecules present after completion = 2 molecules NH3 + 0 molecules O2
+ 8 molecules NO + 12 molecules H2O = 22 molecules.