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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

A mixture of PC15 (12.41 g) and excess NH4Cl was heated at 145 °C for 6 hours. The two reacted in equimolar amounts and evolved 5.14 L of HCl (at STP). Three substances (A, B, and C) were isolated from the reaction mixture. The three substances had the same elemental composition but differed in their molar mass. Substance A had a molar mass of 347.7 g/mol and B had a molar mass of 463.5 g/mol. Give the empirical and molecular formulas for A and B and draw a reasonable Lewis structure for A.

Interpretation Introduction

Interpretation: To give the empirical and molecular formula for A and B and draw the Lewis structure for A.

Concept introduction:

The reaction of PCl5 and NH4Cl results in evolution of hydrochloric acid. The mixture has to be heated at a temperature of 145 °C for six hours. The products formed are found to have the same elemental composition but different molar masses.

Empirical formula of a compound is the smallest integer ratio of numbers of each element presented in that compound.

Molecular formula of a compound is integer multiple of empirical formula, the integer is depend upon the mass of empirical formula and the molecular mass of the compound.

Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.

It is also known as Lewis dot structures which represent the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.

Explanation

The reaction of PCl5 and NH4Cl results in evolution of hydrochloric acid. The balanced reaction is,

    PCl5+NH4Cl4HCl+PNCl2

Thus, the compound A and B have the empirical formula of PNCl2.

It is given that substance A has a molar mass of 347.7 g/mol.

Substance B has a molar mass of 463.5 g/mol.

The molecular formula for PNCl2 is 115.88 g/mol.

Thus, the value of n for A in (PNCl2)n is,

n=347

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Chapter 21 Solutions

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Sect-21.11 P-2.4ACPSect-21.11 P-2.5ACPCh-21 P-1PSCh-21 P-2PSCh-21 P-3PSCh-21 P-4PSCh-21 P-5PSCh-21 P-6PSCh-21 P-7PSCh-21 P-8PSCh-21 P-9PSCh-21 P-10PSCh-21 P-11PSCh-21 P-12PSCh-21 P-13PSCh-21 P-14PSCh-21 P-15PSCh-21 P-16PSCh-21 P-17PSCh-21 P-18PSCh-21 P-19PSCh-21 P-20PSCh-21 P-21PSCh-21 P-22PSCh-21 P-23PSCh-21 P-24PSCh-21 P-25PSCh-21 P-26PSCh-21 P-27PSCh-21 P-28PSCh-21 P-29PSCh-21 P-30PSCh-21 P-31PSCh-21 P-32PSCh-21 P-33PSCh-21 P-34PSCh-21 P-35PSCh-21 P-36PSCh-21 P-37PSCh-21 P-38PSCh-21 P-39PSCh-21 P-40PSCh-21 P-41PSCh-21 P-42PSCh-21 P-43PSCh-21 P-44PSCh-21 P-45PSCh-21 P-46PSCh-21 P-47PSCh-21 P-48PSCh-21 P-49PSCh-21 P-50PSCh-21 P-51PSCh-21 P-52PSCh-21 P-53PSCh-21 P-54PSCh-21 P-55PSCh-21 P-56PSCh-21 P-57PSCh-21 P-58PSCh-21 P-59PSCh-21 P-60PSCh-21 P-61PSCh-21 P-62PSCh-21 P-63PSCh-21 P-64PSCh-21 P-65PSCh-21 P-66PSCh-21 P-67PSCh-21 P-68PSCh-21 P-69PSCh-21 P-70PSCh-21 P-71PSCh-21 P-72PSCh-21 P-73PSCh-21 P-74PSCh-21 P-75PSCh-21 P-76PSCh-21 P-77PSCh-21 P-78PSCh-21 P-79PSCh-21 P-80PSCh-21 P-81PSCh-21 P-82PSCh-21 P-83PSCh-21 P-84PSCh-21 P-85PSCh-21 P-86PSCh-21 P-87PSCh-21 P-88PSCh-21 P-89GQCh-21 P-90GQCh-21 P-91GQCh-21 P-92GQCh-21 P-93GQCh-21 P-94GQCh-21 P-95GQCh-21 P-96GQCh-21 P-97GQCh-21 P-98GQCh-21 P-99GQCh-21 P-100GQCh-21 P-101GQCh-21 P-102GQCh-21 P-103GQCh-21 P-105GQCh-21 P-106GQCh-21 P-107GQCh-21 P-108GQCh-21 P-110GQCh-21 P-111GQCh-21 P-112GQCh-21 P-113GQCh-21 P-114GQCh-21 P-115ILCh-21 P-116ILCh-21 P-117ILCh-21 P-118ILCh-21 P-119ILCh-21 P-120ILCh-21 P-121SCQCh-21 P-122SCQCh-21 P-123SCQCh-21 P-124SCQCh-21 P-125SCQCh-21 P-126SCQCh-21 P-127SCQCh-21 P-128SCQCh-21 P-129SCQCh-21 P-130SCQCh-21 P-131SCQ

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