Question

Una dintre metodele de obtinere a poliacrilonitrilului foloseste ca materii prime metanul, amonicacul si oxigenul. Calculati volumul (c.n.) de metan de puritate 98% necesar obtinerii a 212 kg de poliacrilonitril, daca randamentul intregului proces este de 75%

Answer 1

CH4 + NH3 + 3/2O2 → HCN + 3H2O

2CH4 --> C2H2 + 3H2

C2H2 + HCN --> CH2=CH(CN)--> [-CH2-CH(CN)-]n (poli acrilonitril)

MC3H3N=53g/mol=53kg/kmol

n=m/M=212/53=4 kmoli C3H3N

3 kmoli CH4..............1 kmol C3H3N
x=12 kmoli.................4 kmoli

MCH4=16g/mol=16kg/kmol

n=m/M==>m=n*M=12*16=192 Kg CH4 (mt)

n=mp/mt*100==>mp=n*mt/100=75*192/100=144 kg CH4

n=m/M=144/16=9 kmoli CH4

n=V/Vmolar==>n*Vmolar=9*22.4=201.6m3 CH4

0.98*201.6=197.568m3 CH4 pur