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Chemistry & Chemical Reactivity

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

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

Methane is burned in a laboratory Bunsen burner to give CO2 and water vapor. Methane gas is supplied to the burner at the rate of 5.0 L/min (at a temperature of 28 °C and a pressure of 773 mm Hg). At what rate must oxygen be supplied to the burner (at a pressure of 742 mm Hg and a temperature of 26 °C)?

Interpretation Introduction

Interpretation:

The rate the oxygen be supplied to the Bunsen burner at given temperature and pressure conditions 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.

Avogadro’s Hypothesis:

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

Explanation

Given:

  Rate of methane = 5L/minTemperature =28oC =273.15+28 = 301.15KPressure = 773 mm Hg =1.02atmRate of oxygen = ?Pressure = 742 mm Hg =0.976atmTemperature =26oC =273.15+26 = 299.15K

The balanced chemical equation for the reaction of methane and O2 is as follows,

  CH4+2O22H2O+CO2

Examining the given chemical reaction it is clear that 1 mole of methane reacts with 2 moles of oxygen and gives rise to H2OandCO2.

The available moles of methane is calculated from the given reaction conditions,

   n=PVRT =(1.02atm)(5L)(0

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Chapter 10 Solutions

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Sect-10.6 P-10.11CYUSect-10.7 P-10.12CYUSect-10.8 P-1.1ACPSect-10.8 P-1.2ACPSect-10.8 P-2.1ACPSect-10.8 P-2.2ACPSect-10.8 P-2.3ACPSect-10.8 P-3.1ACPSect-10.8 P-3.2ACPCh-10 P-1PSCh-10 P-2PSCh-10 P-3PSCh-10 P-4PSCh-10 P-5PSCh-10 P-6PSCh-10 P-7PSCh-10 P-8PSCh-10 P-9PSCh-10 P-10PSCh-10 P-11PSCh-10 P-12PSCh-10 P-13PSCh-10 P-14PSCh-10 P-15PSCh-10 P-16PSCh-10 P-17PSCh-10 P-18PSCh-10 P-19PSCh-10 P-20PSCh-10 P-21PSCh-10 P-22PSCh-10 P-23PSCh-10 P-24PSCh-10 P-25PSCh-10 P-26PSCh-10 P-27PSCh-10 P-28PSCh-10 P-29PSCh-10 P-30PSCh-10 P-31PSCh-10 P-32PSCh-10 P-33PSCh-10 P-34PSCh-10 P-35PSCh-10 P-36PSCh-10 P-37PSCh-10 P-38PSCh-10 P-39PSCh-10 P-40PSCh-10 P-41PSCh-10 P-42PSCh-10 P-43PSCh-10 P-44PSCh-10 P-45PSCh-10 P-46PSCh-10 P-47PSCh-10 P-48PSCh-10 P-49PSCh-10 P-50PSCh-10 P-51PSCh-10 P-52PSCh-10 P-53PSCh-10 P-54PSCh-10 P-55PSCh-10 P-56PSCh-10 P-57GQCh-10 P-58GQCh-10 P-59GQCh-10 P-60GQCh-10 P-61GQCh-10 P-62GQCh-10 P-63GQCh-10 P-64GQCh-10 P-65GQCh-10 P-66GQCh-10 P-67GQCh-10 P-68GQCh-10 P-69GQCh-10 P-70GQCh-10 P-71GQCh-10 P-72GQCh-10 P-73GQCh-10 P-74GQCh-10 P-75GQCh-10 P-76GQCh-10 P-77GQCh-10 P-78GQCh-10 P-79GQCh-10 P-80GQCh-10 P-81GQCh-10 P-83GQCh-10 P-84GQCh-10 P-85GQCh-10 P-86GQCh-10 P-87GQCh-10 P-88GQCh-10 P-89GQCh-10 P-90GQCh-10 P-91GQCh-10 P-92GQCh-10 P-93GQCh-10 P-94GQCh-10 P-95ILCh-10 P-96ILCh-10 P-97ILCh-10 P-98ILCh-10 P-99ILCh-10 P-100ILCh-10 P-101ILCh-10 P-102ILCh-10 P-103ILCh-10 P-105ILCh-10 P-106ILCh-10 P-107SCQCh-10 P-108SCQCh-10 P-109SCQCh-10 P-110SCQCh-10 P-111SCQCh-10 P-112SCQCh-10 P-113SCQCh-10 P-114SCQCh-10 P-115SCQCh-10 P-116SCQ

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