Question

14)Prin nitrarea fenolului se formeaza un produs ce contine 15,21 % N.Masa de fenol de puritate 50% necesara obtinerii a 552 kg de produs de nitrare,daca randamentul deste de 75%,este:d)752kg

Answer 1

deaorece nu cunoastem daca e nitrare, dinitrare sau trinitrare, voi scrie ecuatia astfel:
C₆H₅-OH + xHNO₃ -> C₆$H_{5-x}$OH(NO₂)x
MC₆$H_{5-x}$OH(NO₂)x=12*6+5-x+16+1+(14+32)*x=72+5-x+17+46x=45x+94 g/mol
45x+94...............100%
14x.....................15,21%
45x+94=(14x*100/15,21)=92x
92x-45x=47x=94 => n=2 => dinitrare
552kg.................75%
ykg...................100% => y=552*100/75=736kg
94kg..............................184kg
C₆H₅-OH + 2HNO₃ -> C₆H₃OH(NO₂)₂ + 2H₂O
z=?................................736kg
z=94*736/184=376kg fenol pur
376kg..........50%
tkg.............100%
t=376*100/50=752kg fenol cu impuritati