a,
$n_{khí} = PV/RT = \frac{252,56.20}{(273+497).0,082} = 80$ mol=> $n_{N_2}Ư = 16$ mol ; $n_{H_2} = 64$ mol
b,
$N_2 + 3H_2 \rightleftharpoons 2NH_3$
bd $16 64$
pư $0,25.16 3.0,25.16 2.0.25.16$
CB $12 52 8$
c,
$n_{khí sau pư} = 12 + 52 + 8 = 72$ mol
=> $p = \frac{72.0,082.(273+497)}{20} = 227,304$ atm