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134)92 g etanol e oxideaza bland.Stiind ca produsul obtinut formeaza prin tratare cu r fehling 214,5g precipitat rosu,cantitatea de alcool neoxidata este :c)23 g

Răspuns :


46g................................1mol
CH3-CH2-OH + [O] -> CH3-CH=O + H2O
y=69g..........................1,5moli
1mol..........................................................143g
CH3-CH=O + 2Cu(OH)2 -> CH3-COOH + Cu2O + 2H2O
x=1,5moli..................................................214,5g
92-69=23g etanol neoxidate