Respuesta :
Answer:
(I). The gases are present at the end of the experiment are O₂ and NO₂
(II). The pressure of O₂ is 0.158 atm.
The pressure of NO₂ is 0.681 atm
The Pressure of NO is zero.
Explanation:
Given that,
Volume of large bulb = 6.00 L
Pressure = 0.850 atm
Volume of small bulb = 1.50 L
Pressure = 2.50 atm
Temperature = 22°C = 295 K
We need to calculate the moles in NO
Using formula of moles
[tex]n=\dfrac{PV}{RT}[/tex]
Put the value into the formula
[tex]n=\dfrac{0.850\times6.00}{0.0821\times295}[/tex]
[tex]n=0.211\ moles[/tex]
We need to calculate the moles in O
Using formula of moles
[tex]n=\dfrac{PV}{RT}[/tex]
Put the value into the formula
[tex]n=\dfrac{2.50\times1.50}{0.0821\times295}[/tex]
[tex]n=0.155\ moles[/tex]
The balance equation for the reaction is
[tex]2NO+O_{2}\Rightarrow 2NO_{2}[/tex]
So, The gases are present at the end of the experiment are O₂ and NO₂
We need to calculate the remaining moles of O₂
Using formula for remaining moles
Moles of O₂ remaining = [tex]0.155-\dfrac{0.211}{2}[/tex]
[tex]Moles\ of \ O_{2}\ remaining =0.049\ moles[/tex]
Moles of NO₂ = 0.211 moles
Total volume [tex]V=V_{l}+V_{s}[/tex]
Put the value into the formula
[tex]V=6.00+1.50[/tex]
[tex]V=7.5\ V[/tex]
(2). If the gas was consumed completely
We need to calculate the pressure of O₂
Using formula of pressure
[tex]P=\dfrac{moles\times R\times T}{V}[/tex]
Put the value into the formula
[tex]P=\dfrac{0.049\times0.0821\times295}{7.5}[/tex]
[tex]P=0.158\ atm[/tex]
We need to calculate the pressure of NO₂
Using formula of pressure
[tex]P=\dfrac{moles\times R\times T}{V}[/tex]
Put the value into the formula
[tex]P=\dfrac{0.211\times0.0821\times295}{7.5}[/tex]
[tex]P=0.681\ atm[/tex]
If the gas was consumed completely
Then, Pressure of NO is zero.
Hence, (I). The gases are present at the end of the experiment are O₂ and NO₂
(II). The pressure of O₂ is 0.158 atm.
The pressure of NO₂ is 0.681 atm
The Pressure of NO is zero.
The gases that remained at the end of the experiment are nitrogen dioxide and nitric oxide. The partial pressure of nitric oxide has been 0.43 atm, nitrogen dioxide has been 1 atm, and oxygen has been 0 atm.
The balanced chemical equation has been given that, 2 moles of NO reacts completely with 1 mole of oxygen.
(1) According to the ideal gas equation:
PV nRT
n = [tex]\rm \dfrac{PV}{RT}[/tex]
- The moles of NO;
NO = [tex]\rm \dfrac{0.850\;atm\;\times\;6\;L}{0.0821\;L.atm/mol.K\;\times\;295\;K}[/tex]
Moles of NO = 0.211 moles
- The moles of Oxygen:
Oxygen = [tex]\rm \dfrac{2.50\;atm\;\times\;1.50\;L}{0.0821\;L.atm/mol.K\;\times\;295\;K}[/tex]
Moles of oxygen = 0.155 moles.
For 2 moles NO, 1 mole oxygen is used.
For 0.211 moles, Oxygen has been 0.422 moles.
The oxygen has been in a lesser quantity than requires, thus it has been the limiting reactant.
The gas that presents at the end of the reaction has been NO.
(2) The pressure of the gas has been calculated as:
After the complete consumption of the gas, NO remaining has been:
2 moles oxygen = 1 moles NO
0.155 moles oxygen = 0.0775 moles NO.
The moles of NO remaining = 0.211 - 0.0775 mol
The moles of NO remaining = 0.1335 mol.
The total volume of the solution will be 6 + 1.5 L = 7.5 L.
The moles of Nitrogen dioxide formed :
1 mole oxygen = 2 moles Nitric oxide
0.155 moles oxygen = 0.31 moles nitric oxide.
The partial pressure can be given by:
P = [tex]\rm \dfrac{nRT}{V}[/tex]
- The pressure of NO:
Pressure of NO = [tex]\rm \dfrac{0.1335\;\times\;0.08206\;\times\;295}{7.5}[/tex]
Pressure of NO = 0.43 atm.
- The pressure of Oxygen = 0 atm.
The pressure of Nitrogen dioxide:
Pressure of nitrogen dioxide = [tex]\rm \dfrac{0.31\;\times\;0.08206\;\times\;295}{7.5}[/tex]
Pressure of nitrogen dioxide = 1.00 atm.
The gases that remained at the end of the experiment are nitrogen dioxide and nitric oxide. The partial pressure of nitric oxide has been 0.43 atm, nitrogen dioxide has been 1 atm, and oxygen has been 0 atm.
For more information about the pressure of gases, refer to the link:
https://brainly.com/question/16957585
