Mass of empty tube after initial heating Mass of tube plus sample before heating

Question

Mass of empty tube after initial heating Mass of tube plus sample before heating Mass of tube plus sample after heating Volume of water displaced Water temperature (degree C) Barometric pressure Mass of oxygen generated Moles of O_2 generated Vapor pressure of water Pressure of O_2 Temperature of O_2(degree K) Calculated volume of O_2 sample M STP? Calculated molar volume of O_2 at STP Average molar volume Percent error in molar volume Moles of O_2 generated Mass of KClO_3 in sample Percent of KClO_3 in sample Average % KCIO_3

Explanation / Answer

                                trial-1                      trial-2

Mass of oxygen generated 9g)             0.301                       0.351

moles= mass/Moecular weight            =0.301/32= 0.009406                   0.351/32= 0.010969

from the reaction 2KClO3---> 2KCl+3O2

3 moles of O2 gives 2 moles of KClO3

0.009406 moles oxygen gives 0.009406*2/3 ( Trial-1)=0.006271                             0.010969*2/3 =0.007313

Mass of KClO3= moles* Molecular weight =0.006271*122.5 =0.768                           0.007313*122.5= 0.896

Mass of sample before heating= 17.192-15.627= 1.565 gm                    17.467- 15.6 =1.867

percent of KClO3 in the sample = 100*0.768/1.565 =49.07                                 100*0.896/1.867=47.99

Average= (49.07+47.99)/2=48.53%

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