Câu 3:
a)
CTPT xủa X là CnH2n+2O
\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\Rightarrow n_{C_nH_{2n+2}O}=\dfrac{0,4}{n}\left(mol\right)\)
=> \(n_{H_2O}=\dfrac{\dfrac{0,4}{n}.\left(2n+2\right)}{2}=\dfrac{0,4}{n}\left(n+1\right)\left(mol\right)\)
Mà \(n_{H_2O}=\dfrac{9}{18}=0,5\left(mol\right)\)
=> n = 4
=> CTPT: C4H10O
b) \(n_{C_4H_{10}O}=\dfrac{0,4}{4}=0,1\left(mol\right)\)
=> m = 0,1.74 = 7,4 (g)
c)
(1) \(CH_3-CH_2-CH_2-CH_2OH\)
(2) \(CH_3-CH_2-CH\left(OH\right)-CH_3\)
(3) \(CH_3-C\left(CH_3\right)\left(OH\right)-CH_3\)
(4) \(CH_3-CH\left(CH_3\right)-CH_2OH\)
(5) \(CH_3-CH_2-CH_2-O-CH_3\)
(6) \(CH_3-CH\left(CH_3\right)-O-CH_3\)
(7) \(CH_3-CH_2-O-CH_2-CH_3\)
d)
X là \(CH_3-C\left(CH_3\right)\left(OH\right)-CH_3\) (2-metylpropan-2-ol)
mng giải hộ em ạ
