ms = 2480 g , c% = 98%
stim ca c% = mdx100/ms
=> md = msxc%/100 = 2480x98/100 = 2430,4 g etandiol
obtinut la 60%
cat s-ar fi obtinut la 100%..?
60% .................... 2430,4 g etandiol
100% ........................... m = 4050,67 g
n moli
CH2=CH2 + Cl2 --CCl4--> Cl-CH2-CH2-Cl
1
4050,67 g
Cl-CH2-CH2-Cl + 2H2O --(HO-)/-2HCl--> HO-CH2-CH2-OH
62
=> n = 1x4050,67/62 = 65,33 moli etena in c.n.
PV = nRT, T = 273+227 = 500 K , P = 2 atm , V = ? L
=> V = nRT/P = 65,33x0,082x500/2
= 1339,3 L etena
