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CH3OH
mC - 12 u
mH - 4* 1u = 4u
mO - 16u
12/4/16 = 3/1/4
C:H:O = 3:1:4
zad 70
mr - 600 g
C% - 11,5 %
ms - 11,5% * 600g / 100% = 69g
Odp. W 600 g wina znajduje się 69 g etanolu.
zad 80
a - C6H13OH + 9 O2 ----> 6 CO2 + 7 H2O
b - 2 C5H11OH + 5 O2 ---> 10C + 12 H2O
c- 2 C4H10O2 + 3 O2 ---> 8C + 10 H2O