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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

What volume (in liters) of O2, measured at standard temperature and pressure, is required to oxidize 0.400 mol of phosphorus (P4)?

P4(s) + 5 O2(g) → P4O10(s)

Interpretation Introduction

Interpretation:

Considering the given chemical reaction the volume of O2 required to oxidize the given 0.400 mole of phosphorus has to be calculated.

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 = molesofgasP = 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,

  P4 = 0.4 mole= 273.15KP = 760mmHg

In order to calculate the volume of O2 required to oxidize the given amount of phosphorus first the moles of O2 should be determined then finally, by applying the ideal gas equation, the volume of O2 could be determined.

By examining the chemical reaction it is obvious that one mole of P4 requires 5 moles of O2 hence 0.4 moles of P4 requires 0.400 mol of P4=5×0.400 mol of O2  that is 2 mol of O2  needed.

Finally, substituting the moles, pressure temperature into the given ideal gas equation the volume of O2 can be determined as follows,

  R = 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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