Question

200g de formol se trateaza separat cu reactivul Tollens si Fehling.
Calculati:masele din acesti reactivi consumate si cantitatea de acid formic sau metanoic rezultat.

Answer 1

CH2O + 2[Ag(NH3)2](OH) ---> HCOOH + 2Ag + 4NH3 + H2O

Ag(+1) +1e ---> Ag(0). ||2
C(0) -2e ---> C(+2). ||1

CH2O + 2Cu(OH)2 ---> HCOOH + Cu2O + 2H2O

C(0) -2e ---> C(+2). ||2||1
2Cu(+2) +2e ---> Cu2(+1). ||2||1

MCH2O = 30g/mol

n=m/M = 200/30 = 6,66 moli CH2O

1 mol formol.......2*159g Tollens
6,66 moli.....................X = 2120g

1 mol CH2O..........2*98g Fehling
6,666 moli............Y = 1306,53g

n CH2O = nCH2O2 = 6,66 moli

m acid = 6,666*46 = 306,63g * 2 = 613,27g