4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

If a container were to have 16 molecules of NH3 and 16 molecules of O2 initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion? (Use only whole numbers to solve.)

Ah, the classic reactant counting game! Let's get started and see how this reaction plays out.

The balanced equation tells us that 4 molecules of NH3 and 5 molecules of O2 react to form 4 molecules of NO and 6 molecules of H2O.

Since we have 16 molecules of NH3 and 16 molecules of O2 initially, we know that the reactants will be fully consumed. So, we should focus on the products.

For every 4 molecules of NH3 that react, we get 4 molecules of NO. Therefore, since we started with 16 molecules of NH3, we would end up with 16/4 = 4 molecules of NO.

Similarly, for every 5 molecules of O2 that react, we get 6 molecules of H2O. So, when we start with 16 molecules of O2, we end up with 6/5 * 16 = 19.2 molecules of H2O. Since we can't have fractions of molecules, let's consider 19 molecules of H2O instead.

Adding up the products, we have 4 molecules of NO and 19 molecules of H2O. Now, let's not forget about the reactants. So, in total, we would have 16 molecules of NH3, 16 molecules of O2, 4 molecules of NO, and 19 molecules of H2O.

Adding all of that up, we find that the total number of molecules in the container after the reaction goes to completion would be 16 + 16 + 4 + 19 = 55 molecules.

So, we have a grand total of 55 molecules, making for a pretty crowded party in that little container!

To determine the number of total molecules present in the container after the reaction goes to completion, we need to calculate the total number of molecules in the reactants (NH3 and O2) and products (NO and H2O).

Given:
- Number of NH3 molecules = 16
- Number of O2 molecules = 16

We can determine the total number of reactant molecules by summing the quantities of NH3 and O2.

Total number of reactant molecules = Number of NH3 molecules + Number of O2 molecules
= 16 + 16
= 32

Now, let's calculate the total number of product molecules.

From the balanced chemical equation:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

The stoichiometry of the balanced chemical equation tells us that for the formation of 4 NO molecules, we need 4 molecules of NH3 and 5 molecules of O2.
Similarly, for the formation of 6 H2O molecules, we also need 4 molecules of NH3 and 5 molecules of O2.

Hence, the number of product molecules will be equal to the reactant molecules.

Total number of product molecules = Total number of reactant molecules
= 32

Therefore, the total number of molecules (reactants plus products) in the container after the reaction goes to completion is 32 molecules.

To determine the total number of molecules present in the container after the reaction goes to completion, we need to calculate the moles of NH3 and O2 initially and then use the stoichiometry of the reaction to find the final numbers of molecules for all the species involved.

1. Start by calculating the moles of NH3 and O2 initially:
Given that there are 16 molecules of NH3 and 16 molecules of O2, we need to convert them to moles. The molar mass of NH3 is 17.031 g/mol, and O2 is 32.00 g/mol.

Moles of NH3 = (16 molecules) / (6.022 × 10^23 molecules/mol) ≈ 2.66 × 10^-24 mol
Moles of O2 = (16 molecules) / (6.022 × 10^23 molecules/mol) ≈ 2.66 × 10^-24 mol

2. Use the balanced equation to determine the stoichiometry of the reaction:
From the balanced equation, we can deduce that 4 moles of NH3 react with 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O.

3. Calculate the number of molecules of NO and H2O produced:
Moles of NO = (4/4) × (2.66 × 10^-24 mol) = 2.66 × 10^-24 mol
Moles of H2O = (6/4) × (2.66 × 10^-24 mol) = 3.99 × 10^-24 mol

4. Convert the moles of NO and H2O to molecules:
Number of molecules of NO = (2.66 × 10^-24 mol) × (6.022 × 10^23 molecules/mol) ≈ 1.60 molecules
Number of molecules of H2O = (3.99 × 10^-24 mol) × (6.022 × 10^23 molecules/mol) ≈ 2.41 molecules

5. Calculate the total number of molecules in the container:
To find the total number of molecules present in the container after the reaction, we sum up the initial molecules of NH3 and O2 with the final molecules of NO and H2O:
Total number of molecules = initial molecules of NH3 + initial molecules of O2 + final molecules of NO + final molecules of H2O
Total number of molecules = 16 + 16 + 1.60 + 2.41 ≈ 36 molecules

Therefore, there would be approximately 36 total molecules present in the container after the reaction goes to completion.

I'm assuming from the statement about whole numbers that we don't "count" parts of a molecule.

What's the limiting reagent(LR)?
16 NH3 x (4 NO/4 NH3) = 16 x 4/4 = 16 molecules NO formed if we had 16 molecules NH3 and all the O2 we needed.

16 O2 x (4 NO/5 O2) = 16 x 4/5 = 12.8 molecules NO formed if we had 16 molecules O2 and all the NH3 we needed.

Therefore, O2 is the LR and NH3 is the excess reagent (ER).

Next we calculate molecules NO and H2O formed from the O2.
For NO:
16 O2 x (4 NO/5 O2) = 12.8 molecules NO formed. Since we are to use ONLY whole numbers, this is 12 molecules NO and we used 15 of the 16 molecules O2 which leaves 1 molecule O2 unused. You can check that out by
15 O2 x 4/5 = 12 molecules NO formed. The 16 O2 - 15 O2 = 1 molecule O2 left unreacted.

For H2O:
We're using only 15 of the O2 now so
15 O2 x (6 H2O/5 O2) = 15 x 6/5 = 18 molecules. That's 18 molecules H2O formed (of course with the 1 molecule O2 left over but that's the same molecule O2 we had left over from the NO calculation).

How much NH3 is used:
15 O2 x (4 NH3/5 O2) = 15 x 4/5 = 12 molecules NH3 used. We had 16 to begin with so we have 4 molecules NH3 left over.
Total molecules in the container now is
12 NO formed
1 O2 unreacted
18 H2O formed
4 NH3 unreacted
total is 12 + 1 + 18 + 4 = ?