Use stoichiometric methods to analyze a mixture.Close ProblemA mixture of MgCl, and inert material is analyzed to determine the Mg content. First the mixture is dissolved in water. Then allof the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate.MgCl2(aq) + 2A£NO3(aq) 2A£CI(s) + Mg(NO3)½(aq)In one experiment, a 0.5770 g sample of the mixture resulted in 1.3717 g of AgCl. Determine the percent (by mass) of Mg in themixture.% MgCheck & Submit AnswerShow ApproachA huluFEBétv12F10F8

Answered: Use stoichiometric methods to analyze a…

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter6: Types Of Chemical Reactions And Solution Stoichiometry

Section: Chapter Questions

Problem 126CP

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Use stoichiometric methods to analyze a mixture.
Close Problem
A mixture of MgCl, and inert material is analyzed to determine the Mg content. First the mixture is dissolved in water. Then all
of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate.
MgCl2(aq) + 2A£NO3(aq) 2A£CI(s) + Mg(NO3)½(aq)
In one experiment, a 0.5770 g sample of the mixture resulted in 1.3717 g of AgCl. Determine the percent (by mass) of Mg in the
mixture.
% Mg
Check & Submit Answer
Show Approach
A hulu
FEB
étv
12
F10
F8

Expert Solution

Step 1

$G i v e n : M a s s o f s a m p l e = 0.5770 g m a s s o f A g C l f o r m e d = 1.3717 g a l s o f r o m t a b l e : m o l a r m a s s o f M g C l_{2} = 95.21 g m o l a r m a s s o f A g C l = 143.32 g g i v e n r e a c t i o n : M g C l_{2} + 2 A g N O_{3} \to 2 A g C l + M g (N O_{3}) 1 m o l e o f M g C l_{2} g i v e s 2 m o l e s o f A g C l n o o f m o l e s o f A g C l f o r m e d = \frac{m a s s o f A g C l f o r m e d}{m o l a r m a s s o f A g C l} = \frac{1.3717 g}{143.32 g} = 0.0096 m o l e A s, 1 m o l e o f M g C l_{2} g i v e s 2 m o l e s o f A g C l i . e n o o f m o l e o f M g C l_{2} c o n s u m e d = 1 / 2 t h e n o o f m o l e o f A g C l f o r m e d ∴ n o o f m o l e o f M g C l_{2} u s e d = \frac{1}{2} x 0.0096 m o l e = 0.0048 m o l e$

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