Question

1.Ce cantitate de acid cianhidric(HCN) se obtine prin amonodirarea a 4480m3 CH4 de puritate 95%, cu un randament de 85%?
2.Ce volum de etalon(C2H5OH) cu densitatea de 0,8g/cm3 se obtine prin reactia cu apa a 448l etana(C2H4) masurati la 4 atm si 00C cu un randament de 90%?

Answer 1

1.
0.95*4480=4256m3 CH4 pur

n=V/Vmolar=4.256/22.4=0.19 kmoli CH4 pur

n=m/M==>n=n*M=0.19*16=3.04kg CH4 (mt)

n=mp/mt*100==>mp=n*mt/100=85*3.04/100=2.584kg CH4


2CH4 + 2NH3 + 3O2 --> 2HCN + 6 H2O

32g CH4...34kg NH3..96kg O2....54kg HCN
2.584kg...x=2.7455kg...y=7.752kg...z=4.3605kg HCN

n=m/M=4.3605/27=0.1615 kmoli HCN.


2. n=V/Vmolar=448/22.4=20 moli C2H4

pV= ??? T°=?