Question

Ce volum de etenă (c.n) este necesar pentru a obține 5.000 kg oxid de etenă de puritate 88% cu un randament de 80%?

A. 1792 m^3
B. 2800 m^3
C. 3500 m^3
D. 2314 m^3
E. 361 m^3

Answer 1

C2H2 + 1/2 O2 = C2H2O
puritatea = m pura/m impura * 100 => m pura = 5000*88/100 = 4400 kg
randament = m practica/m teoretica * 100 
=> m teoretica = 4400/80 * 100 = 5500 kg 
22,4 l C2H2..........44 g etoxid
x m3 C2H2............5500 kg etoxid
x = 2800 m3

Answer 2

CH2=CH2 + 1/2O2 ---> CH2(-O-)CH2

puritatea=masa pura / masa impura *100 ==> m pur=p*mimpur/100=88*5/100=4.4 tone = 4.400 kg oxid de etena

n(eta)=m practic/m teoretic * 100 ==> mt=mp*100/n=4400*100/80=5500 kg oxid de etena

MC2H4=28kg/kmol
MC2H4O=44kg/kmol

28 kg CH2=CH2 (etena)......................44 kg C2H4O (etoxid)
x=28*5500/44...............................................5500 kg
x=3500 kg etena

n=m/M=3500/28=125 kmoli

n=V/Vm==>V=n*Vm=125*22.4=2800 m³ CH2=CH2