Question

Se fabrica 6,75 tone de formol 40% din alcool metilic cu un randament de 90%. Cantitatea de alcool metilic de concentratie 98% folosita este....

Answer 1

Reactie: CH3-OH -> CH2=O + H2
c/100=md/ms => md=c*ms/100=6,75*40/100=2,7t acetaldehida
90/100=mpractica/mteoretica=2,7/mteoretica => mteoretica=2,7*100/90=3t acetaldehida=3000kg acetaldehida
32kg.............30kg
CH3-OH -> CH2=O + H2
x=?..............3000kg
x=3000*32/30=3200kg metanol
98/100=3200/ms => ms=3200*100/98=3265,3kg=3,2653t solutie alcool metilic 98%