Problem 6.144 A 3.444 g sample of an unknown alkaline-earth metal was allowed to
ID: 510275 • Letter: P
Question
Problem 6.144
A 3.444 g sample of an unknown alkaline-earth metal was allowed to react with a volume of chlorine gas that contains 1.91×1022Cl2 molecules. The resulting metal chloride was analyzed for chlorine by dissolving a 0.818 g sample in water and adding an excess of AgNO3(aq) to give a precipitate of 1.126 g of solid AgCl.
Part A
What is the percent Cl in the alkaline-earth chloride?
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Part B
What is the identity of the alkaline-earth metal?
Express your answer as a chemical symbol.
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%Explanation / Answer
MCl2 + 2 AgNO3 ----------------> 2 AgCl + M (NO3)2
moles of Cl = moles of AgNO3
= 1.126 / 143.32
= 0.007856
mass of Cl = moles x molar mass
= 0.007856 x 35.45
= 0.27854 g
% mass of Cl = (mass of Cl / mass of MCl2 sample ) x 100
= (0.27854 / 0.818 ) x 100
% mass of Cl = 34.1 %
the metal is Mg