trioleina + 3H2 => tristearina
m triol. = 1768 g=> n triol. = 1768 g : 885, 4 g/mol= aprox.2 mol(1.99 mol)
1 mol trioleina consuma 3 mol H2
2 mol trioleina consuma x mol H2
x= 2 x 3 /1 = 6 mol.
pV=nRT
V=nRT/p => V= 6 x 0.082 x 350 /3.5=172.2/3.5= 49,2 l H2
n=6 mol
R=0.082 l x atm/ mol x K
T= 273 + 77= 350 K
p=3.5 atm