Respuesta :
Answer:
64 molecules
Explanation:
we have the reaction
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
That is a balanced equation
We can see here that we need 5 molecules of O2 for every 4 molecules of NH3
we have in the beginning 10 molecules of O2 and 52 molecules of NH3
we can say:
if 4 molecules of NH3 need 5 molecules of O2 how many molecules are needed for 52 molecules of NH3
?
4 molecules of NH3 →5 molecules of O2
52 molecules of NH3 →X molecules of O2
[tex]X=\frac{52 molecules-of- NH_3* 5 molecules-of-O_2}{4 molecules-of-NH_3}=65 molecules-of- O_2[/tex]
Since we just have 10 molecules of O2 it is not enough so this is the limiting reagent and all the calculations have to be done base on it.
We use simple rule of three to get all the molecules.
5 molecules of O_2 → 4 molecules of NH_3
10 molecules of O_2 →X molecules of NH_3
[tex]\frac{4 molecules-of- NH_3* 10 molecules-of-O_2}{5 molecules-of-O_2}=8 molecules-of- NH_3[/tex]
Which means that we use all the O2 but just 8 molecules of NH3 and we have 44 less.
NO produced
5 molecules of O_2 → 4 molecules of NO
10 molecules of O_2 → X molecules of NO
[tex]\frac{4 molecules-of- NO* 10 molecules-of-O_2}{5 molecules-of-O_2}=8 molecules-of- NO[/tex]
H2O produced
5 molecules of O_2 → 6 molecules of H_2 O
10 molecules of O_2 → X molecules ofH_2 O
[tex]\frac{4 molecules-of- H_2 O* 10 molecules-of-O_2}{5 molecules-of-O_2}=12 molecules-of- H_2 O[/tex]
Total molecules
44 molecules of NH3 + 8 molecules of NO + 12 molecules of H2O = 64 molecules.
64 molecules will be present after the reaction is completed.
The reaction of any compounds and molecules occurs in the presence of the reactants and produces products through various means.
How to calculate the number of molecules?
The reaction can be shown as,
[tex]\rm 4NH_{3}(g) + 5O_{2}(g) \rightarrow 4NO(g) + 6H_{2}O(g)[/tex]
From the equation, it can be said that,
5 molecules of oxygen are present for 4 molecules of ammonia.
If the container has initially, 10 molecules of oxygen and 52 molecules of ammonia then,
4 molecules of [tex]\rm NH_{3}[/tex] reacts with 5 molecules of [tex]\rm O_{2}[/tex]
52 molecules of [tex]\rm NH_{3[/tex] will react with X molecules of [tex]\rm O_{2}[/tex]
So solving for X we get,
[tex]\begin{aligned}\rm X &= \dfrac{52 \times 5}{4}\\\\&= 65 \;\rm molecules \;of \rm \;O_{2}\end{aligned}[/tex]
From this, it can be said that the oxygen is the limiting reagent and the further calculation will be:
5 molecules of [tex]\rm O_{2}[/tex] reacts with 4 molecules of [tex]\rm NH_{3}[/tex]
10 molecules [tex]\rm O_{2}[/tex] react with X molecules of [tex]\rm NH_{3}[/tex]
Solving further X for Ammonia:
[tex]\begin{aligned}\rm X &= \dfrac{4 \times 10}{5}\\\\&= 8 \;\rm molecules \;of\; \rm NH_{3}\end{aligned}[/tex]
Now calculate the number of Nitric oxides produced:
5 molecules of [tex]\rm O_{2}[/tex] = 4 molecules of [tex]\rm NO[/tex]
10 molecules of [tex]\rm O_{2}[/tex] = X molecules of [tex]\rm NO[/tex]
Solving X for nitric oxide:
[tex]\begin{aligned}\rm X &= \dfrac{4 \times 10}{5}\\\\&= 8 \rm \; molecules \;of \rm \;NO\end{aligned}[/tex]
Now calculate the number of water produced:
5 molecules of [tex]\rm O_{2}[/tex] = 6 molecules of [tex]\rm H_{2}O[/tex]
10 molecules of [tex]\rm O_{2}[/tex] = X molecules of [tex]\rm H_{2}O[/tex]
Solving X for water:
[tex]\begin{aligned}\rm X &= \dfrac{4 \times 10}{5}\\\\&= 8 \;\rm molecules\; of\; \rm H_{2}O\end{aligned}[/tex]
Total molecules will be calculated as:
[tex]\rm 44 \;molecules \;of\; NH_{3} + 8 \;molecules \;of \rm \;NO + 12 \;molecules \;of \;H_{2}O = 64 \;\rm molecules[/tex]
Therefore, 64 molecules are produced.
Learn more about reactants and products here:
https://brainly.com/question/16742690
