Trial 1 Trial 2 5 5q Mass of empty tube after initial heating Mass of tube plus

Question

Trial 1 Trial 2 5 5q 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 (C) Barometric pressure 58 1130 tat_-117g- CALCULATIONS Show your caleulations in the space provided on the next page Molar Volume of Oxygen: Mass of oxygen generated Moles of O, generated Vapor pressure of water Pressure of O Temperature of O, ('K) Calculated volume of O, sample at STP Calculated molar volume of O, at ST Average molar volume Percent error in molar volume

Explanation / Answer

2KClO3(s) --->2KCl(s)+3O2(g)

Barometric reading=742.50 torr

Patm=PO2+PH2O [PH2O=vapour pressure of water at 24 degrees]

PO2=Patm-PH2O=742.50torr-22.4 torr=720.1 torr

Volume of O2 at STP=V(STP)

V(STP)*P(STP)/T(STP)=PO2*V/T

VSTP=V(T(STP)/T) *(PO2/P(STP))=

VO2=145ml+147ml/2=146ml (average of trial 1 and trial 2)

T=273.15+24=297.15K

V(STP)=146ml*273.15K*720.1torr/297.15K*760torr=127.162 ml

weight of O2 produced=weight of tube before rxn-weight of tube after rxn=Average(17.2+17.8/2) - av(16.958+17.220/2)=17.5-17.089=0.411g

mol of O2 =0.411g*32g/mol=0.0128mol

molar volume of O2=V=RT/P=127.162 ml*1L/1000ml/0.0128mol=9.934L/mol

mol KClO3 decomposed=0.0128mol O2*(2mol KClO3/3mol O2)=0.00853 mol

weight of KClO3 decomposed=0.00853mol*122.51g/mol=1.045g

Percent KClO3=Actual wt of KClO3/Initial wt of sample

Initial wt of sample=mass of sample before heating-mass of tube

mass of sample before heating=15.587+15.596/2=15.591g

Initial wt of sample=17.5-15.591=1.908

percent KClO3=1.045g/1.908g*100=54.77%

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