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?
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.
Oh, let me calculate that for you! So, according to the balanced equation you provided, for every 4 molecules of NH3 and 5 molecules of O2, you'll get 4 molecules of NO and 6 molecules of H2O.
Starting with 10 molecules of NH3 and 10 molecules of O2, if the reaction goes to completion, you would end up with 10 molecules of NO and 15 molecules of H2O. So, in total, that would be 10 + 10 + 10 + 15 = 45 molecules in the container.
That's quite a molecular party going on in there!
To determine the total number of molecules present in the container after the reaction goes to completion, we need to calculate the number of molecules for each reactant and product and then find their sum.
Let's analyze the balanced chemical equation:
4NH3 + 5O2 → 4NO + 6H2O
From the equation, we can deduce the following ratios:
- 4 moles of NH3 react with 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O.
Since we are given the initial quantities in terms of molecules, we'll convert the 10 molecules of NH3 and O2 into moles to determine the limiting reactant.
To convert from molecules to moles, we need to use Avogadro's number, which states that 1 mole is equal to 6.022 × 10^23 molecules.
For NH3:
10 molecules NH3 * (1 mole NH3 / 6.022 × 10^23 molecules NH3) = 1.66 × 10^-23 moles NH3
For O2:
10 molecules O2 * (1 mole O2 / 6.022 × 10^23 molecules O2) = 1.66 × 10^-23 moles O2
Since the ratio of NH3 to O2 in the equation is 4:5, we can see that O2 is the limiting reactant because we have an equal number of molecules of NH3 and O2 initially. Thus, all the O2 will be consumed in the reaction.
Now let's find the number of moles for all the products using the balanced equation:
Moles of NO = 4 moles NO * (1.66 × 10^-23 moles NO / 1.66 × 10^-23 moles O2) = 4 moles NO
Moles of H2O = 6 moles H2O * (1.66 × 10^-23 moles H2O / 1.66 × 10^-23 moles O2) = 6 moles H2O
Since each mole contains 6.022 × 10^23 molecules, we need to multiply the number of moles by Avogadro's number to find the number of molecules for each product:
Molecules of NO = 4 moles NO * (6.022 × 10^23 molecules / 1 mole) = 2.409 × 10^24 molecules NO
Molecules of H2O = 6 moles H2O * (6.022 × 10^23 molecules / 1 mole) = 3.613 × 10^24 molecules H2O
Finally, to calculate the total number of molecules (reactants plus products) in the container after the reaction goes to completion, we add the molecules of reactants and products:
Total molecules = Molecules of NH3 + Molecules of O2 + Molecules of NO + Molecules of H2O
= 10 molecules + 10 molecules + 2.409 × 10^24 molecules + 3.613 × 10^24 molecules
= 6.022 × 10^23 + 6.022 × 10^23 + 2.409 × 10^24 + 3.613 × 10^24
= 6.614 × 10^24 molecules
So, after the reaction goes to completion, there would be approximately 6.614 × 10^24 molecules in the container.
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