Two bulbs are connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.750 atm, and the small bulb, with a volume of 1.50 L, contains oxygen at a pressure of 2.50 atm. The temperature at the beginning and the end of the experiment is 22 ∘ C . After the stopcock is opened, the gases mix and react. 2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g )

Determine which gases remain after the reaction goes to completion and calculate their partial pressures.

The temperature at the beginning and the end of the experiment is 22 degree celsius.

Which gases are present at the end of the experiment?

What are the partial pressures of the gases?

If the gas was consumed completely, put 0 for the answer.

Partial pressure of O2 = 0.200 atm

Partial pressure of NO2 = 0.600 atm

O2 and NO2 remain at the end

Explanation:

Step 1: Data given

Volume of the large bulb (nitric oxide) = 6.00 L

Pressure = 0.750 atm

Volume of the small bulb (oxygen) = 1.50 L

Pressure = 2.50 atm

Temperature = 22.0 °C = 295 K

Step 2: The balanced equation

2NO(g) + O2(g) → 2NO2(g)

Step 3: Calculate moles

p*V = n*R*T

n= (p*V)/(R*T)

⇒with n = the moles of NO= TO BE DETERMINED

⇒with p = the pressure of NO = 0.750 atm

⇒with V = the volume of NO = 6.00 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 295 K

n = (0.750 * 6.00) /( 0.08206*295)

n = 0.186 moles

p*V = n*R*T

n= (p*V)/(R*T)

⇒with n = the moles of O2 = TO BE DETERMINED

⇒with p = the pressure of O2 = 2.50 atm

⇒with V = the volume of O2 = 1.50 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 295 K

n = (2.50 * 1.50) /( 0.08206*295)

n = 0.155 moles

Step 4: Calculate the limiting reactant

For 2 moles NO we need 1 mol O2 to produe 2 moles NO2

No is the limiting reactant. It will completely be consumed (0.186 moles). O2 is in excess. There will react 0.186/ 2 =  0.093 moles

There will remain 0.155 - 0.093 = 0.062 moles

Step 5: Calculate moles NO2

For 2 moles NO we need 1 mol O2 to produe 2 moles NO2

For 0.186 moles NO we'll have 0.186 moles NO2

After completion only O2 and NO2 will remain

Step 6: Calculate the partial pressure after the reaction

p*V = n*R*T

p= (n*R*T)/V

⇒with n = the moles of O2 = 0.062 moles

⇒with p = the pressure of O2 = TO BE DETERMINED

⇒with V = the volume of O2 =7.50 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 295 K

p = (0.062 * 0.08206 * 295) / 7.50

p = 0.200 atm

p= (n*R*T)/V

⇒with n = the moles of NO2 = 0.186 moles

⇒with p = the pressure of NO2 = TO BE DETERMINED

⇒with V = the volume of NO2 =7.50 L

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 295 K

p = (0.186*0.08206 * 295) / 7.50

p = 0.600 atm

the answer to your question is c.

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