The picture below shows two bulbs connected by a stopcock. The large bulb, with

Question

The picture below shows two bulbs connected by a stopcock. The large bulb, with a volume of 6.00 L, contains nitric oxide at a pressure of 0.800 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 degree C. After the stopcock is opened, the gases mix and react: 2NO(g) + O_2(g) rightarrow 2NO_2(g) Which gases are present at the end of the experiment? What are the partial pressures of the gases?

Explanation / Answer

2 NO(g) + O2(g) ---> 2NO2(g)

No of mol of NO = PV/RT = 6*0.8/(0.0821*295) = 0.198 mol

No of mol of O2 = 1.5*2.5/(0.0821*295) = 0.155 mol

from equation,

2 mol NO = 1 mol O2

limiting reagent is NO

so that , after the reaction NO2,O2 is present.

No of mol of NO2 produced   = PV/RT = 6*0.8/(0.0821*295) = 0.198 mol

Partial pressure of NO = 0 atm

Partial pressure of NO2 = (0.198*0.198*0.0821*295)/(6+1.5)

                      = 0.64 atm

Partial pressure of o2 = ((0.155-0.099)*0.198*0.0821*295)/(6+1.5)

= 0.036 atm

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