The picture below shows two bulbs connected by a stopcock. The 6.00L bulb contai

Question

The picture below shows two bulbs connected by a stopcock. The 6.00L bulb contains nitric oxide at a pressure of 0.500 atm, and the 1.50-L. bulb contains oxygen at a pressure of 2.50 atm. After the stopcock is opened, the gases mix and react: 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 C Which gases are present at the end at the experiment? What are the partial pressures of the gases? If the gas was consumed completely, put 0 for the answer

Explanation / Answer

Gas constant R=.0821 L ATM K-1 mol-1

Before reaction,

moles of no = Nno=P*V/R*T=.5*6/.0821*298=.124 mol

moles of o2 = No2=2.5*1.5/(.0821*298)=.153 mol

After reaction at constant temperature

2 moles of no requires 1 mole of o2 to give 2 moles of no2

.124 moles no react with.124/2=.062 mole of o2 to give .124 moles of no2

So all no will be completely consumed

After reaction,

moles of no, Nno=0,

moles of o2, No2=.153-.062=.091 mole,

moles of no2, Nno2 =.124 moles

Total no of molecules = nt=.091+.124=.215 moles

mole fraction:

for no, Xno=0 as no is limiting reactant, so all no consumed

for no2, Xno2=Nno2/nt=.124/.215=.577

for o2, Xo2= No2/nt=.091/.215=.423

Total volume Vt=6+1.5=7.5L

Total pressure=Pt=nt*R*T/V=.215*.0821*298/7.5=.7 atm

partial pressure of o2= Xo2*Pt=.423*.7=0.3 atm

partial pressure of no2= Xno2*Pt=.577*.7=.4 atm

partial pressure of no= 0 atm

At the end of experiment, o2 and no2 gases are present.

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