Đáp án đúng:
Giải chi tiết:\({C_n}{H_{2n + 1}}C{H_2}OH{\rm{ }} + {\rm{ }}{O_2}\buildrel {xt} \over \longrightarrow {C_n}{H_{2n + 1}}COOH{\rm{ }} + {\rm{ }}{H_2}O\)
Giả sử X có: x (mol) → (14n + 32)x = 9,6 (1)
\(\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,{C_n}{H_{2n + 1}}C{H_2}OH{\rm{ }} + {\rm{ }}{O_2}\buildrel {xt} \over \longrightarrow {C_n}{H_{2n + 1}}COOH{\rm{ }} + {\rm{ }}{H_2}O \cr & pu:\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a \cr & du:\,\,\,\,\,\,\,\,\,\,x - a \cr} \)
-OH + Na → -ONa + ½ H2
(x + a) → 0,5(x + a)→ 0,5.(x + a) = 0,24 (2)
Từ (1) và (2) \(\left\{ \matrix{\left( {14n{\rm{ + }}\,32} \right)x{\rm{ = }}9,6 \hfill \cr x{\rm{ + }}\,a{\rm{ = }}\,0,48\buildrel {x > a} \over\longrightarrow 0,24 < x < 0,48 \hfill \cr} \right. = > n = 1 = > C{H_3}OH\)
=> nCH3OH ban đầu = 0,3 và sau phản ứng = 0,18
=> H = 60%