Question

690 de grame triol cu raport de masa carbon la hidrogen la oxigen 9 la 2 la 12 se incalzeste in prezenta de h2so4 concentrat dand produsul b . 2/3 din b reactioneatza cu hidrogen in prezenta de nichel si formeaza x grame compus c. x este?"

Answer 1

mC:mH:mO=9:2:12
fiind triol => 3 atomi de O => mO=16*3=48
mC:mH:mO=9:2:12|•4=36:8:48
nr.at.C=mC/MC=36/12=3
nr.at.H=mH/MH=8/1=8
nr.at.O=mO/MO=48/16=3
f.m: C3H8O3 => HO-CH2-CH(OH)-CH2-OH
92g............................................1mol
HO-CH2-CH(OH)-CH2-OH -> CH2=CH-CH=O + 2H2O
690g..........................................a=?
a=690/92=7,5moli
66% din 7,5=66*7,5/100=5moli
1mol....................................60g
CH2=CH-CH=O + 2H2 -> CH3-CH2-CH2-OH
5moli...................................x=300g