Bài 5 trang 144 - Sách giáo khoa Hoá 9
\(CH_2 = CH_2 + H_2O \xrightarrow[]{xúc tác} CH_3-CH_2-OH\)
1mol 46g
H = \(\dfrac{13,8}{46}\)x 100% = 30%
\(CH_2 = CH_2 + H_2O \xrightarrow[]{xúc tác} CH_3-CH_2-OH\)
1mol 46g
H = \(\dfrac{13,8}{46}\)x 100% = 30%