what volume of O2 is measured at 24 degrees Celsius and 0.630 atm will be produced by the decomposition of 3.00 g KClO3? Assuming 100% yield. (R=0.08206 L atm/mol K)

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molar mass of KClO3= 122.5495

moles of KClO3 IN 3 gm=3.0/122.5495=0.0245 molesfrom the equation, moles of O2=3/2 *0.0245=0.03675from PV=nRTV=nRT/P ;T=24+273=297 =0.03675*0.08206* 297 /0.630 =1.42168 LITRES

moles of KClO3 IN 3 gm=3.0/122.5495=0.0245 moles

from the equation, moles of O2=3/2 *0.0245=0.03675

from PV=nRT

V=nRT/P ;T=24+273=297

=0.03675*0.08206* 297 /0.630

=1.42168 LITRES

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