## Chemistry and Chemical Reactivity (9th Edition)

a) The number of moles of Ammonia is: $562\ g\div17.03\ g/mol=33.0\ mol$ From stoichiometry: $33×3/2=49.5\ mol\ H_2$ $PV=nRT$ $V=49.5\ mol×0.082×(56+273)\ K/(745/760×1\ atm)$ $V=1362.3\ L$ b) Number of moles of nitrogen: $33.0×1/2=16.5\ mol$ Number of moles of air: $16.5/x=78.1/100$ $x=21.13\ mol$ $V=21.13\ mol×0.082×(29+273)\ K/(745/760×1\ atm)$ $V=534.1\ atm$