## Chemistry and Chemical Reactivity (9th Edition)

a) Number of moles of He: $PV=nRT$ $n=1\ atm×12.5\ L/0.082057\ L.atm/K.mol/(21.5+273)\ K$ $n=0.517\ mol$ Mass: $0.517\ mol×4.0026\ g/mol=2.07\ g$ b,c d) Since only volume and number of moles change: $V/n=constant$ $12.5\ L/0.517\ mol=26\ L/n_f$ $n_f=1.08\ mol$ From Dalton's law: He: $0.517/1.08×100\%=48\ mol\%$ $O_2$: $(1.08-0.517)/1.08×100\%=52\ mol\%$ $P_{He}=1\ atm×48/100=0.48\ atm$ $P_{O_2}=1\ atm×52/100=0.52\ atm$