Respuesta :
Answer:
0.02 moles of O₂ will be leftover.
Explanation:
The reaction is:
2NO(g) + O₂(g) → 2NO₂(g) (1)
We have the mass of NO and O₂, so we need to find the number of moles:
[tex] n_{NO} = \frac{m}{M} = \frac{24.7 g}{30.01 g/mol} = 0.82 moles [/tex]
[tex] n_{O_{2}} = \frac{m}{M} = \frac{13.8 g}{31.99 g/mol} = 0.43 moles [/tex]
From equation (1) we have that 2 moles of NO reacts with 1 mol of O₂ to produce 2 moles of NO₂, so the excess reactant is:
[tex] n_{NO} = \frac{2}{1}*0.43 moles = 0.86 moles [/tex]
[tex]n_{O_{2}} = \frac{1}{2}*0.82 moles = 0.41 moles[/tex]
Hence, from above we can see that the excess reactant is O₂ since 0.41 moles react with 0.86 moles of NO and we have 0.43 moles in total for O₂.
The number of moles of excess reactant is:
[tex]n_{T} = 0.43 moles - 0.41 moles = 0.02 moles[/tex]
Therefore, 0.02 moles of O₂ will be leftover.
I hope it helps you!
The number of moles of excess reactant that would be left over is 0.0197 mole
From the question,
We are to determine the number of moles of excess reactant that would be left over.
The given balanced chemical equation for the reaction is
2NO(g) + O₂(g) → 2NO₂ (g)
This means,
2 moles of NO is needed to completely react with 1 mole of O₂
Now, we will determine the number of moles of each reactant present
- For NO
Mass = 24.7 g
Molar mass = 30.01 g/mol
From the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
∴ Number of moles of NO present = [tex]\frac{24.7}{30.01}[/tex]
Number of moles of NO present = 0.823059 mole
- For O₂
Mass = 13.8 g
Molar mass = 32.0 g/mol
∴ Number of moles of O₂ present = [tex]\frac{13.8}{32.0}[/tex]
Number of moles of O₂ present = 0.43125 mole
Since,
2 moles of NO is needed to completely react with 1 mole of O₂
Then,
0.823059 mole of NO is will react with completely react with [tex]\frac{0.823059 }{2}[/tex] mole of O₂
[tex]\frac{0.823059 }{2} = 0.4115295[/tex]
∴ Number of moles of O₂ that reacted is 0.4115295 mole
This means O₂ is the excess reactant and NO is the limiting reactant
Now, for the number of moles of excess reactant left over
Number of moles of excess reactant left over = Number of moles of O₂ present - Number of moles of O₂ that reacted
∴ Number of moles of excess reactant left over = 0.43125 mole - 0.4115295 mole
Number of moles of excess reactant left over = 0.0197205 mole
Number of moles of excess reactant left over ≅ 0.0197 mole
Hence, the number of moles of excess reactant that would be left over is 0.0197 mole
Learn more here: https://brainly.com/question/18757768
