## Chemistry and Chemical Reactivity (9th Edition)

Number of moles of octane: $0.048\ g\div 114.23\ g/mol=4.20×10^{-4}\ mol$ From stoichiometry: $4.20×10^{-4}\ mol×18/2=0.00378\ mol$ of water produced $4.20×10^{-4}\ mol×25/2=0.00525\ mol$ of oxygen required From ideal gas law: $P.V=n.R.T$ Water vapor $P=0.00378\ mol×0.082057\ L.atm/mol.K×(30+273)\ K/4.75\ L$ $P=0.0198\ atm$ Oxygen $P=0.00525\ mol×0.082057\ L.atm/mol.K×(22+273)\ K/4.75\ L$ $P=0.0268\ atm$