... CH3CH2OH 43,39%; Y: CH3CH2CH2OH ; Z: CH3CHOHCH3 B X: CH3CH2OH 33,39%; Y: CH3CH2CH2OH Z: CH3CHOHCH3 C X: CH3CH2CH2OH 43,39%; Y: CH3CH2CH2CH2OH Z: CH3CH2CHOHCH3 D X: CH3CH2CH2OH 33,39%; Y: CH3CH2CH2CH2OH ... - mH2 → mH2 = 15,6 - 15 ,2 = 0,4 gam → nH2 = 0 ,2 mol → nR-OH = 0,4 mol o H SO4 d , 140 C • 2R-OH → R2O + H2O Ta có: nH2O = 0,4 : = 0 ,2 mol Theo BTKL: mete = 15,6 - 0 ,2 x 18 = 12 gam Câu 23 : ... có CTC CnH2n + 1OH o H SO4 d , 140 C 2CnH2n + 1OH → (CnH2n + 1)2O + H2O Theo BTKL: mH2O = 21 0,8 - 183,3 = 27 ,5 gam → nH2O ≈ 1, 528 mol → nA = 1, 528 x = 3,056 mol → MCnH2n + 1OH = 21 0,8 : 3,056...