Question

In the first trial, the % KClO3 in mixture should equal 11, my calculation is a little bit off, I can't figure out why? and how? Trial #2 calculation is fine though.

Answer 1

Second calculations are not correct since the percentage of KClO3 mentioned is not 11%, but 110%. (100*2.451/2.161)= 113% ( The calculations need to be verified since the mixture mass is less than the mass of KClO3 in the sample.

Mass of empty test tube + mixture= 22.363 g

Mass of test tube = 19.918 gm

Mass of mixture= 22.363-19.918=2.445 gms

Mass of test tube + residue= 22.312 gms

Mass of oxygen = mass of test tube+mixture- mass of test tube+residue =

Moles of oxygen= 0.051/32=1.6*10-3 moles ( that escaped )

From the reaction 2KClO3-à 2KCl +3O2

Moles of KClO3= 1.6*10-3*2/3=0.001067

Mass of KClO3= 0.001067* molecular weight of KClO3= 0.001067*122.5=0.1307 gms

Mass % KClO3= 100*0.1307/2.445=5.345% is correct as per the data furnished.