MOLAR VOLUME OF A GAS DATA Before reaction: a. unknown number weight of beaker,

Question

MOLAR VOLUME OF A GAS DATA Before reaction: a. unknown number weight of beaker, test tube b. c weight of beaker, test tube, KCIo3-KCI d. weight of beaker, test tube, KCIo3-KCI. Mno2- 01o e. weight of KCIO3 KCI sample (c After reaction: f. volume of oxygen collected at laboratory LG2.r (determined temperature and pressure by the amount of water expelled from the flask) g. temperature of the oxygen h. water vapor pressure at the measured temperature (table: Handbook of Chemistry and Physics) i. barometric pressure j. pressure of oxygen collected li-h) k. weight of beaker, test tube, Mno2, residue 1. weight of oxygen released during reaction (d -ky CALCULATIONS I. Using the following steps with the proper number of significant figures, calculate the molar volume of oxygen showing all work clearly. A. the moles of oxygen gas formed (use mass of gas from

Explanation / Answer

from mass,

weight of O2=46.21g

moles of O2=46.21g/32g/mol=1.444 moles

so using balanced eqn

2KClO3 +MnO2------->2KCl+MnO2+3O2

2 moles KClO3=3 moles O2

so, 3/2 moles O2 is generated from 1 moles KClO3=

3/2 *1.444=2.166 moles*122.55g/mol=265.443g

% KClO3=mass of KClO3 reacted/total mass of KCLO3-KCl*100

from volume,

PV=nRT

n=PV/RT

P=745.303 mmHg /760mmHg/atm=0.98 atm

n=0.98 atm*33.026 L/0.0821 Latm/K.mol*298K=1.323 moles

moles of O2=1.323 moles

mass of O2=1.323 moles*32g/mol=42.332 g

moles of KClO3 reacted=3/2O2=3/2 *1.323=1.984 moles

mass of KClO3 reacted=1.984 moles*122.55 g/mol=243.200 g

%KClO3==mass of KClO3 reacted/total mass of KCLO3-KCl*100

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