A test tube containing a mixture of KCl, KClO3, and MnO3 having a total weight o

Question

A test tube containing a mixture of KCl, KClO3, and MnO3 having a total weight of 23.584g was heated to decompose the KClO3. After heating the mass was found to be 22.347 g.
A) how many moles of oxygen gas were evolved?
B) calculate the number of moles and the mass of KClO3 originally present in the test tube
C) if the original mass of the sample was 4.39 g calculate the precent KClO3 in the original sample.

NOTE: the numbers next to the mixture are suppose to be lowered under them not next to them

Explanation / Answer

A) The reaction for the decomposition of KClO3 is,

2KClO3 ----> 2KCl + 3O2

We have mixture of KCl, KClO3 + MnO3 weight = 23.584 g

After heating weight becomes = 22.347 g

Reduction in mass = 23.584 - 22.347 = 1.237 g

mass of O2 = 1.237 g

moles of O2 gas evolved = mass/molar mass = 1.237/32 = 0.039 mole

B) Moles of KClO3 in the sample = 2 x 0.039/3 = 0.026 mole

mass of KClO3 = 0.026 x 122.55 = 3.19 g

C) If the original sample weight = 4.39 g

KClO3 in sample = 4.39 x 3.19/22.347 = 0.627 g

% of KClO3 = 0.627/4.39 = 14.28%

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