## Chemistry and Chemical Reactivity (9th Edition)

Number of moles of oxygen: $n=PV/RT$ $n=(735/760×1\ atm)×0.327\ L/((0.082)(19+273)\ K)$ $n=0.0132\ mol$ From stoichiometry, the number of moles of $KClO_3$ is: $0.0132\ mol×2/3=0.0088\ mol$ This has a mass of: $122.55\ g/mol×0.0088\ mol=1.08\ g$ Mass fraction: $1.08\div1.56×100\%=69\%$