$n_{KMnO_4}=\dfrac{25,28.25\%}{158}=0,04(mol)$
Đặt CTTQ anken X là $C_nH_{2n}$
$3C_nH_{2n}+2KMnO_4+4H_2O\to 3C_nH_{2n}(OH)_2+2MnO_2+2KOH$
$\to n_X=\dfrac{3}{2}n_{KMnO_4}=0,06(mol)$
$M_X=\dfrac{4,2}{0,06}=70=14n$
$\to n=5$
Vậy CTPT X là $C_5H_{10}$
Các CTCT:
$CH_2=CH-CH_2-CH_2-CH_3$
$CH_3-CH=CH-CH_2-CH_3$
$CH_2=C(CH_3)-CH_2-CH_3$
$CH_3-C(CH_3)=CH-CH_3$
$CH_3-CH(CH_3)-CH=CH_2$
