Ancol

G

gacon_lonton_timban


n(H2O) = 0,3 mol = n(3 ete )
n(mỗi ete) = 0,3 : 3 = 0,1 mol
2ROH ---> ROR + H2O
2R'OH ----> R'OR' + H2O
ROH + R'OH ---> ROR' + H2O
=> 0,1 ( 3R + 3R' + 16.3 ) = 22,1
=> R + R' = 58
=> Ch3OH , C3H7OH


 
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