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composition of potassium chlorate questions and problems
REPORT FOR EXPERIMENT 9 (continued) B. Qualitative Examination of Residue 1. Record what you observed when silver nitrate was
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Answer #1

Answer: Part B.

1. a) When silver nitrate was added to potassium chloride, a white precipitate of silver chloride is formed which in insoluble in water.

AgNO3 + KCl -> AgCl (white ppt) +KNO3

b) When silver nitrate is reacted with potassium chlorate, no white precipitate of AgCl is formed

AgNO3 + KClO3 -> Ag(ClO3) + KNO3

c and 2 a) When silver nitrate is added to the residue then also we observed formation of white precipitate which indicates formation of AgCl. Therefore, it can be concluded that the residue contains potassium chloride.

Answer 2 b) If silver nitrate is added to the solution of sodium chloride then also the reaction occurs which leads to the formation of white AgCl which is insoluble in water.

AgNO3 + NaCl -> AgCl + NaNO3

Answer 2 c) The residue is actually potassium chloride as it forms a white precipitate on reaction with silver nitrate. If the residue is potassium chlorate then the reaction doesn't occur to furnish AgCl. The formation of silver chloride is a clear proof that the residue contain potassium chloride.

Questions and Problems

Answer 1: If the student put potassium chloride instead of potassium chlorate the reaction which is suppose to be oxidation turns out to be reduction as KCl is reducing agent. Therefore, it the student is trying to oxidize something, it undergoes reduction. The crucible when heated with potassium chloride in it would result in melting of the species.

Answer 2: The % oxygen in experiment is lower as there is no oxygen in KCl and due to contamination of potassium chlorate with KCl there will be less KClO3 in experiment.

Answer 3: If potassium chlorate sample is contaminated with moisture then there will be higher experimental % oxygen as there is more oxygen in sample.

Answer 4: Molecular weight of all elements

Al = 27

Cl = 35.5 (3 x Cl = 106.5)

O = 16 (9 x O = 144)

Total mass 27 + 106.5 +144 = 277.5

Cl = 106.5/277.5 x 100 = 38.37 %

Answer 5:

Ca(ClO3)2 ->  CaCl2 + 3O2

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