   Chapter 9, Problem 65QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
8 views

# Although they were formerly called the inert gases, at least the heavier elements of Group 8 do form relatively stable compounds. For example, xenon combines directly with elemental fluorine at elevated temperatures in the presence of a nickel catalyst.:math> Xe ( g ) + 2 E F 2 ( g ) → XeF 4 ( s ) at is the theoretical mass of xenon tetrafluoride that should form when 130. g of xenon is reacted with 100. g of F 2 ? What is the percent yield if only 145 g of XeF 4 is actually isolated?

Interpretation Introduction

Interpretation:

To determine the theoretical yield and percentage yield of XeF4.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

The limiting reactant in a particular reaction has due to following properties:

1. Limiting reactant completely reacted in a particular reaction.
2. Limiting reactant determines the amount of the product in mole.
Explanation

Theoretical yield is also known as calculated yield. Theoretical yield is the amount of product which is theatrically calculates by the amount of limiting agent. It is the maximum amount of product which is formed in any reaction.

Other name of actual yield is observed yield.

The percentage yield is calculated by the actual or observed experimental yield divided by the theoretical yield and multiplied by 100.

The percentage yield can be calculates by the use of following expression:

Actual yield or given yield Theoretical yield or calculated yield ×100%

Theoretical yield is the greatest amount of a product possible from a reaction, when assume that the reaction goes to 100% competition.

The balance equation for the calcium and oxalate ions is as follows:

Xe(g)+2F2(g)  XeF4(s)

Given:

Amount of Xe = 130. g

Amount of F2 = 100. g

Calculation:

Number of moles of Xe and F2 are calculated as follows:

Number of moles=mass in gmolarmass=130. g131.29  g/mol=0.990 moles XeNumber of moles=mass in gmolarmass=100. g37

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