Chapter 10, Problem 95IL

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# You have a 550.-mL tank of gas with a pressure of 1.56 atm at 24 °C. You thought the gas was pure carbon monoxide gas, CO, but you later found it was contaminated by small quantities of gaseous CO2 and O2. Analysis shows that the tank pressure is 1.34 atm (at 24 °C) if the CO2 is removed. Another experiment shows that 0.0870 g of O2 can be removed chemically. What are the masses of CO and CO2 in the tank, and what is the partial pressure of each of the three gases at 25 °C?

Interpretation Introduction

Interpretation:

Considering the given chemical reaction under given temperature and pressure conditions the mass of CO2 and CO and the partial pressure of each of the given gases should be determined.

Concept Introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained.  It is referred as ideal gas equation.

nTPV = RnTPPV = nRTwhere,n = moles of gasP = pressureT = temperatureR = gas constant

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties. At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to its volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

Explanation

Given:

â€‚Â VolumeÂ ofÂ tankÂ =Â 550mL=550Ã—10-3LPressureâ€Šâ€Š=â€Šâ€Š1.56Â atmTemperatureÂ =Â 24oCâ€Šâ€Š=â€Šâ€Š273.15+24â€Šâ€Š=â€Šâ€Š297.15KPÂ atÂ RemovalÂ ofÂ CO2â€Šâ€Š=â€Š1.34atmTatÂ RemovalÂ ofÂ CO2â€Šâ€Š=24oCâ€Šâ€Š=â€Šâ€Š273.15+24â€Šâ€Š=â€Šâ€Š297.15KMassÂ ofÂ removedÂ O2=0.0870g

The pressure of CO2 is found to be 0.22atm since it is given that the original pressure for the tank is 1.56Â atm when CO2 is removed the pressure is found to be 1.34atm hence both the values are to be subtracted to gain the pressure for CO2.

Now, the moles of CO2 is calculated from the given set of conditions,

â€‚Â PV=nRTn=â€Šâ€ŠPVRTâ€Šâ€Š=â€Šâ€Šâ€Š0.22Ã—550Ã—10âˆ’30.0821Ã—297.15=4.96Ã—10âˆ’3moles

â€‚Â moles=massMolarâ€Šâ€Šmassmass=molesÃ—Molarâ€Šâ€Šmassâ€Šâ€Š=â€Šâ€Š4.96Ã—10âˆ’3Ã—44.01g/molâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=â€Š0.218g

The total number of moles present in the given chemical reaction is found to be

â€‚Â PVÂ =Â nRTnÂ =PVRTâ€Šâ€Š=â€Šâ€Š1.56Ã—0.5500.0821Ã—297.15â€Šâ€Š=â€Šâ€Š0.0352moles

The given mass of oxygen is converted into moles as follows,

â€‚Â MolesÂ ofÂ O2â€Šâ€Š=â€Šâ€ŠmassMolarÂ massâ€Šâ€Š=â€Šâ€Š0.087032g/mol=2.72Ã—10âˆ’3g

Now, the moles of carbon monoxide is obtained as follows,

â€‚Â TotalÂ molesÂ =Â MolesÂ ofÂ O2+MolesÂ ofÂ CO2+Â MolesÂ ofÂ COMolesÂ ofÂ COâ€Šâ€Š=â€Šâ€ŠTotalÂ molesâˆ’(MolesÂ ofÂ O2+MolesÂ ofÂ CO2)â€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=â€Šâ€Š0.0352âˆ’(2.72Ã—10âˆ’3+4.96Ã—10âˆ’3)â€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š=â€Šâ€Š0

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