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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

Ni(CO)4 can be made by reacting finely divided nickel with gaseous CO. If you have CO in a 1.50-L flask at a pressure of 418 mm Hg at 25.0 °C, along with 0.450 g of Ni powder, what is the theoretical yield of Ni(CO)4?

Interpretation Introduction

Interpretation:

The treatment of given amount of Ni with carbon monoxide under given temperature and pressure conditions gives rise to Ni(CO)4 and the theoretical yield for that product formed should be determined.

Concept introduction:

Ideal gas Equation:

Any gas can 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,

  Ni = 0.450 gpressure, P =418 mm Hg=418 mm Hg760 mm Hg=0.55atmtemperature, T = 25oC=273.15+25=298.15KNi(CO)4=?V = 1.50L

In order to calculate the theoretical yield for the given product first the balanced chemical reaction should be determined.

The balanced chemical reaction for the given reaction is as follows,

  Ni+4CONi(CO)4

  moleNi=givenmassmolarmass=0.4558.7=7.67×103

  moleofCOPV=nRTn=PVRT=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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