Question

a 550 ml sample b of oxygen gas is collected over water at a temperature of 21.0°C and a pressure of 643 mmHg according to the following reaction 2KCLO3(S)   2KCL(S) + 3O2(g)

Answer 1

Antoine equation for vapor pressure of water is log psat(mm Hg)= 8.07131-1730.63/(t+233.426), t in deg.c

Vapor pressure of water at 21 deg.c= 8.07131-1730.63/(21+233.426)= 1.269, Psat (mm Hg)= 18.59 mm Hg

partial pressure of Oxygen, P = total pressure- vapor pressrue of water= 643-18.59= 624.41 mm Hg

760mm Hg= 1atm, P= 624.41/760 atm =0.8215 atm, V= volume of gas= 550ml, 1000ml= 1L, V= 550/1000=0.55L

T=21deg.c= 21+273= 294K, R= gas constant= 0.0821 L.atm/mole.K

from gas law PV= nRT, n= moles of gas= PV/RT= 0.8215*0.55/(0.0821*294) =0.0187 moles

mass of oxygen gas= moles of oxygen gas collected* molar mass= 0.0187*32= 0.5984 gm

from the reaction, 3 moles of oxygen is generated from 2 moles of KClO3

0.0187 moles of oxygen is generated from 2*0.0187/3 moles of KClO3= 0.0125 moles

molar mass of KClO3=122.5, mass of KCLO3 used= 0.0125* 122.5 =1.53125 gm