Asked by harley
A 0.6753 g sample of an unknown metal was
converted to the nitrate, MCl2, then a solution of the
nitrate was treated with silver nitrate to give
MCl2 (aq) + 2 AgNO3 → M(NO3)2 + 2 AgCl(s)
silver chloride crystals. The silver chloride was
isolated and weighed 1.7220 g. What was the metal?
I've have worked it out and gotten 41.55 whcih is Ca, but the prof says the correct answer is Cd, which is 112.41.
converted to the nitrate, MCl2, then a solution of the
nitrate was treated with silver nitrate to give
MCl2 (aq) + 2 AgNO3 → M(NO3)2 + 2 AgCl(s)
silver chloride crystals. The silver chloride was
isolated and weighed 1.7220 g. What was the metal?
I've have worked it out and gotten 41.55 whcih is Ca, but the prof says the correct answer is Cd, which is 112.41.
Answers
Answered by
DrBob222
The story you tell of the metal is not very good; somehow the nitrate and chloride became mixed up. At any rate
M --> M(NO3)2--> MCl2 --> 2AgCl
mols AgCl = 1.722/143.3209 = 0.01201.
moles M = 1/2 that.
0.01201/2 = 0.006 but you need to do it more accurately.
molar mass = grams/mol = 0.6753/0006 = about 112
M --> M(NO3)2--> MCl2 --> 2AgCl
mols AgCl = 1.722/143.3209 = 0.01201.
moles M = 1/2 that.
0.01201/2 = 0.006 but you need to do it more accurately.
molar mass = grams/mol = 0.6753/0006 = about 112
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