Chapter 10, Problem 70GQ

### 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

# Propane reacts with oxygen to give carbon dioxide and water vapor.C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)If you mix C3H8 and O2 in the correct stoichiometric ratio, and if the total pressure of the mixture is 288 mm Hg, what are the partial pressures of C3H8 and O2? If the temperature and volume do not change, what is the pressure of the water vapor after reaction?

Interpretation Introduction

Interpretation:

For the given chemical reaction with total pressure of the mixture the partial pressure of gases C3H8andO2 should be determined and also that with same temperature and volume the pressure of the water vapor after reaction should be determined.

Concept introduction:

Ideal gas Equation:

Any gas can be 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.

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Partial pressure: The partial pressure for any gas can be obtained by multiplication of total pressure of the gas with the mole fraction of the gas present in that total mixture.

Mole fraction: It defines the amount of particular species present in the mixture.  It is obtained by dividing the mole of gas by the total mole of gas present in the mixture.

Explanation

Given,

â€‚Â TotalÂ PressureÂ ofÂ theÂ mixtureÂ Ptotal=Â 288Â mmÂ HgPartialÂ pressureÂ forÂ C3H8â€Šâ€Š=â€Šâ€Š?PartialÂ pressureÂ forÂ O2â€Šâ€Š=â€Šâ€Š?

The given balanced equation for the combustion of given compound in order to obtain the given set of gas is as follows,

â€‚Â C3H8(g)+5O2(g)â†’3CO2(g)+4H2O(g)

Therefore, the total mole of gases produced from the combustion of given compound is 7.

Total mole of gases used in the reaction is about 6 moles.

The partial pressure for a gas is determined by multiplication of total pressure with the mole fraction of that gas present in the total amount.

â€‚Â PC3H8=â€Šâ€ŠXC3H8Ã—Ptotalâ€Šâ€Š=â€Šâ€Šâ€Š16Ã—288â€Šâ€ŠmmHg=48â€Šâ€ŠmmHgPO2=â€Šâ€ŠXO2Ã—Ptotalâ€Šâ€Š=â€Šâ€Šâ€Š56Ã—288â€Šâ€ŠmmHg=240â€Šâ€ŠmmHg

Hence, the partial pressure of individual gases was determined.

After completion of the reaction there are totally about 7 moles of gases produced that is 3 moles of CO2 and 4 moles of H2O from the reaction.

That is for given reaction,

â€‚Â 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€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€ŠC3H8(g)+5O2(g)â€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ†’â€Šâ€Šâ€Šâ€Šâ€Šâ€Š3CO2(g)+â€Šâ€Šâ€Šâ€Šâ€Š4H2O(g)Initialâ€Šâ€ŠPressureâ€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€Š48â€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Šâ€Š

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