The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.500 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. what are the partial gasses of No, and No2?

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Chemistry

6 November, 06:31

The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.500 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. what are the partial gasses of No, and No2?

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Ciara · Answer 1

P (O2) = 0.300 atm

P (NO2) = 0.400 atm

P (NO) = 0 atm

Explanation:

Step 1: Data given

Volume of large bulb = 6.00L

Large bulb contains nitric oxide at 0.500 atm

Volume of the small bulb = 1.50 L

Small bulb contains oxygen at 2/50 atm

The initial temperature = 22.0°C

Step 2: The balanced equation

2 NO + O2 → 2 NO2

Step 3: Calculate moles of NO

P*V=n*R*T

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

⇒ with P = the pressure of NO = 0.500 atm

⇒ with V = the volume of NO = 6.00L

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

⇒ with T = the temperature = 22.0 °C = 295 K

⇒ with n = the number of moles of NO

n (NO) = (0.500 * 6.00) / (0.08206*295)

n (NO) = 0.124 moles

Step 4: Calculate moles of O2

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

⇒ with P = the pressure of O2 = 2.50 atm

⇒ with V = the volume of 02 = 1.50 L

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

⇒ with T = the temperature = 22.0 °C = 295 K

⇒ with n = the number of moles of O2

n (O2) = (2.50*1.50) / (0.08206*295)

n (O2) = 0.155 moles

Step 5: Calculate the limiting reactant

For 2 moles NO consumed, we need 1 mole O2 to produce 2 moles of NO2

No is the limiting reactant. It will completely be consumed. (0.124 moles)

O2 is in excess. There will react 0.124/2 = 0.062 moles O2

There will remain 0.155 - 0.062 = 0.093 moles O2

Step 6: Calculate moles of NO2

For 2 moles NO consumed, we need 1 mole O2 to produce 2 moles of NO2

For 0.124 moles NO, we'll have 0.124 moles NO2

Step 7: Calculate partial pressure of O2

P*V = n*R*T

P = (n*R*T) / V

⇒ n = the number of moles O2 = 0.093 moles O2 remain

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

⇒ T = the temperature = 295 Kelvin

⇒ V = the volume = 6.00 + 1.50 = 7.50 L

P (O2) = 0.300 atm

Step 8: Calculate partial pressure of NO2

P = (n*R*T) / V

⇒ n = the number of moles NO2 = 0.124 moles

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

⇒ T = the temperature = 295 Kelvin

⇒ V = the volume = 6.00 + 1.50 = 7.50 L

P (NO2) = 0.400 atm

P (NO) = 0 atm because NO is completely consumed.