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?

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.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

c. electrical charges of ionic bonds.