V = 448 L, P = 3atm , t = 127oC
=> T = 273+127 = 400 K
PV = nRT => n = PV/RT
= 3x448/0,082x400 = 40,98 moli propena
40,98 moli m g
C3H6 + H2O --H+--> C3H7OH
1 60
=> m = 40,98x60/1 = 2458,54 g alcool s-a obtinut teoretic.... dara la noi reactia a decurs doar 60%
=> 100% ........................ 2458,54 g
60% ........................ md = 1475,12 g alcool
stim ca c% = mdx100/ms
=> ms = mdx100/c% = 1475,12x100/80 = 1843,9 g sol. de propanol
