H CHO +H2-->CH3OH xa moli
CH3-CHO+H2--->CH3-CHOH x b moli
-CALCULEZ moli H2: niu= 8,96l/22,4l/mol= >0,4molH2
a +b=0,4
ax(2gH+12gC+16gO) +bx(24gC+4gH+16gO)= 14,8
Rezolva sistemul
Total moli aldehide-->0,4
0,4.moli..........a,moli metanal........b,moli etanal
10............ x..................................y
CH3OH +O---> CH2O+H2O
1mol metanol....1mol metanal....
o,4mol.............o,4mol........... ---> m= 0,4mol x32g/mol metanol