Bài 4 trang 119 Tài liệu Dạy - học Hoá học 9 tập 1

Đề bài

Viết phương trình hoá học thực hiện chuỗi biến hoá sau:

$S\buildrel {\left( 1 \right)} \over \longrightarrow {H_2}S\buildrel {\left( 2 \right)} \over \longrightarrow S{O_2}\buildrel {\left( 3 \right)} \over \longrightarrow S{O_3}\buildrel {\left( 4 \right)} \over \longrightarrow$$\, {H_2}S{O_4}\buildrel {\left( 5 \right)} \over \longrightarrow CuS{O_4}\buildrel {\left( 6 \right)} \over \longrightarrow Cu{\left( {OH} \right)_2}$$\,\buildrel {\left( 7 \right)} \over \longrightarrow CuO$

Lời giải chi tiết

$\eqalign{ & (1)\,\,\,S\,\,\, + \,\,\,{H_2}\,\,\xrightarrow{{{t^o}}}\,\,\,{H_2}S \cr & (2)\,\,\,2{H_2}S\,\,\,\, + \,\,\,\,3{O_2}\,\xrightarrow{{{t^o}}}\,\,\,2S{O_2}\,\, + \,\,\,2{H_2}O \cr & (3)\,\,\,2S{O_2}\,\,\, + \,\,\,{O_2}\,\,\,\, \overset {{t^o},xt} \leftrightarrows  \,\,\,2S{O_3} \cr & (4)\,\,\,S{O_3}\,\, + \,\,\,{H_2}O\,\,\, \to \,\,\,{H_2}S{O_4} \cr & (5)\,\,\,{H_2}S{O_4}\,\, + \,\,\,CuO\,\,\, \to \,\,\,CuS{O_4}\,\, + \,\,{H_2}O \cr & (6)\,\,\,CuS{O_4}\,\,\, + \,\,\,\,2NaOH\,\, \to \,\,\,Cu{(OH)_2} \downarrow \,\, + \,\,\,N{a_2}S{O_4} \cr & (7)\,\,Cu{(OH)_2}\,\xrightarrow{{{t^o}}}\,\,CuO\,\, + \,\,{H_2}O \cr}$

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