Đáp án:
a) ${n_{{N_2}(du)}} = 25,92mol$; ${n_{{H_2}(du)}} = 77,76mol$; ${n_{N{H_3}}} = 12,96mol$
b) $334,75atm$
Giải thích các bước giải:
a)
$pV = nRT \Rightarrow {n_{hh(trc)}} = \dfrac{{372.20}}{{0,082.(427 + 273)}} = 129,6mol$
${n_{{N_2}}}:{n_{{H_2}}} = 1:3 \\\Rightarrow {n_{{N_2}}} = 32,4mol;{n_{{H_2}}} = 97,2mol$
PTHH: ${N_2} + 3{H_2}\overset {} \leftrightarrows 2N{H_3}$
$H = 20\%$
$→{n_{{N_2}(pư)}} = 32,4.20\% = 6,48mol\\ \Rightarrow {n_{{N_2}(dư)}} = 25,92mol$
${n_{{H_2}(pư)}} = 97,2.20\% = 19,44mol \\\Rightarrow {n_{{H_2}(dư)}} = 77,76mol$
${n_{N{H_3}}} = 2{n_{{N_2}(pư)}} = 12,96mol$
b)
${n_{hhsau}} = 25,92 + 77,76 + 12,96 = 116,64mol$
$ \Rightarrow p = \dfrac{{nRT}}{V} = \dfrac{{116,64.0,082.(427 + 273)}}{{20}} = 334,75atm$