   Chapter 10, Problem 95IL

Chapter
Section
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:

Volume of tank = 550mL=550×10-3LPressure=1.56 atmTemperature = 24oC=273.15+24=297.15KP at Removal of CO2=1.34atmTat Removal of CO2=24oC=273.15+24=297.15KMass of removed 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 1.56 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,

PV=nRTn=PVRT=0.22×550×1030.0821×297.15=4.96×103moles

moles=massMolarmassmass=moles×Molarmass=4.96×103×44.01g/mol=0.218g

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

PV = nRTn =PVRT=1.56×0.5500.0821×297.15=0.0352moles

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

Moles of O2=massMolar mass=0.087032g/mol=2.72×103g

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

Total moles = Moles of O2+Moles of CO2+ Moles of COMoles of CO=Total moles(Moles of O2+Moles of CO2)=0.0352(2.72×103+4.96×103)=0

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 