Asked by Micheal

35.8g of mixture of kcl and kclo3,were heated to constant to a constant mass.if the residue weighed 24.9g,what was the percentage mass of the kcl int the mixture.kcl is not decomposed onm heating
Answered by Micheal
Help me out
Answered by Micheal
Solve it for me please
Answered by DrBob222
mass KClO3 + mass KCl before heating = 35.8 grams
mass KCl after heating to constant weight = 24.9 grams.
35.8 g - 24.9 =
2KClO3 ==> 2KCl + 3O2.
Answered by DrBob222
oops. pushed the wrong button and it posted too early.
mass KClO3 + mass KCl before heating = 35.8 grams
mass KCl after heating to constant weight = 24.9 grams.
35.8 g - 24.9 = 10.9 g = mass lost due to oxygen escaping.
2KClO3 ==> 2KCl + 3O2.
mols O2 lost = 10.9 g/32 = 0.341
mols KClO3 that decomposed = 0.341 x (2 mols KClO3/3 mols O2) = 0.341 x 2/3 = ?
grams KClO3 = moles KClO3 x molar mass KClO3 = ?
mass mixture - mass KClO3 = mass KCl
% KCl in the original mixture = (grams kcl from above/mass sample)100 = ?
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