## Chemistry and Chemical Reactivity (9th Edition)

Number of moles of diborane: $P.V=n.R.T$ $n=0.15\ atm×1.5\ L/0.082057\ L.atm/K.mol/(25+273)\ K$ $n=9.20×10^{-3}\ mol$ From stoichiometry, number of moles of $O_2$ $9.20×10^{-3}\ mol×3/1=0.0276\ mol$ Mass: $0.0276\ mol×32.0\ g/mol=0.88\ g$