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. MgCly(aq) + 2A9NO3(aq) 2AGCI(s) + Mg(NO3)2(aq) In one experiment, a 0.5770 g sample of the mixture resulted in 1.3717 g of AgCI. Determine the percent (by mass) of Mg in the mixture. % Mg
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. MgCly(aq) + 2A9NO3(aq) 2AGCI(s) + Mg(NO3)2(aq) In one experiment, a 0.5770 g sample of the mixture resulted in 1.3717 g of AgCI. Determine the percent (by mass) of Mg in the mixture. % Mg
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 108AE
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Transcribed Image Text:Use stoichiometric methods to analyze a mixture.
A mixture of MGCI, 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.
MgCl,(aq) + 2A£NO3(aq) 2A9CI(s) + Mg(NO3)2(aq)
In one experiment, a 0.5770 g sample of the mixture resulted in 1.3717 g of AgCI. Determine the percent (by mass) of Mg in the
mixture.
% Mg
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Expert Solution
Step 1: Interpretation
The given reaction is a precipitation reaction in which the precipitate formed is AgCl.
Steps involved to determine % mass of Mg are:
Step 1:
we should determine number of moles of AgCl precipitated.
Step 2:
From number of moles of AgCl precipitated, number of moles of Mg can be interpreted stoichiometrically.
Step 3:
Mass of Mg can de determined from number of moles of Mg by substituting in mole formula.
Step 4:
%mass of Mg in mixture is obtained by substituting the mass of Mg in %mass formula.
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