20 kg ulei cu 100% acid oleic
a kg .................. 84,6% => a = 16,92 kg
16,92 kg n kmoli
CH3-(CH2)7-CH=CH-(CH2)7-COOH + H2 => CH3-(CH2)16-COOH
282 1
=> n = 16,92x1/282 = 0,06 kmoli = 60 moli H2
stim ca PV = nRT
=> V = nRT/P = 60x0,082x(273+150)/3
= 693,72 L H2