   Chapter 3, Problem 187CP

Chapter
Section
Textbook Problem

Lanthanum was reacted with hydrogen in a given experiment to produce the nonstoichiomctric compound LaH2.90. Assuming that the compound contains H−, La2+, and La3−, calculate the fractions of La2+ and La3+ present.

Interpretation Introduction

Interpretation:

The mole fractions of La2+ and La3+ ions in La2.90H should be determined.

Concept introduction:

Mole fraction:

The ratio between amount of the analyte present in sample and total amount of sample to give a mole fraction of the analyte present in taken sample.

Mole fraction=AnalyteamountinsampleTotalamountofsample

Explanation

To determine the mole fractions of La2+and La3+ions in La2.90H

The ionic compounds are having electrical neutrality by making number of cations and anions with respect with their charges.

The total charge of an ionic compound is given by sum of multiplications of charge and coefficients of all the ions present in the compound.

From the above statement, La2.90H is contains La2+, La3+ and H- ions.

(n×La2++m×La3+)+(-1)H=0n+m=2.902×n+3×m=2.902×0

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