Answer
See the answer below.
Work Step by Step
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:
$16.5/x=78.1/100$
$x=21.13\ mol$
$V=21.13\ mol×0.082×(29+273)\ K/(745/760×1\ atm)$
$V=534.1\ atm$
