Viết phương trình hoá học thực hiện chuỗi biến hoá sau:
\(S\buildrel {\left( 1 \right)} \over
\longrightarrow {H_2}S\buildrel {\left( 2 \right)} \over
\longrightarrow S{O_2}\buildrel {\left( 3 \right)} \over
\longrightarrow S{O_3}\buildrel {\left( 4 \right)} \over
\longrightarrow\)\(\, {H_2}S{O_4}\buildrel {\left( 5 \right)} \over
\longrightarrow CuS{O_4}\buildrel {\left( 6 \right)} \over
\longrightarrow Cu{\left( {OH} \right)_2}\)\(\,\buildrel {\left( 7 \right)} \over
\longrightarrow CuO\)
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\(\eqalign{
& (1)\,\,\,S\,\,\, + \,\,\,{H_2}\,\,\xrightarrow{{{t^o}}}\,\,\,{H_2}S \cr
& (2)\,\,\,2{H_2}S\,\,\,\, + \,\,\,\,3{O_2}\,\xrightarrow{{{t^o}}}\,\,\,2S{O_2}\,\, + \,\,\,2{H_2}O \cr
& (3)\,\,\,2S{O_2}\,\,\, + \,\,\,{O_2}\,\,\,\, \overset {{t^o},xt} \leftrightarrows \,\,\,2S{O_3} \cr
& (4)\,\,\,S{O_3}\,\, + \,\,\,{H_2}O\,\,\, \to \,\,\,{H_2}S{O_4} \cr
& (5)\,\,\,{H_2}S{O_4}\,\, + \,\,\,CuO\,\,\, \to \,\,\,CuS{O_4}\,\, + \,\,{H_2}O \cr
& (6)\,\,\,CuS{O_4}\,\,\, + \,\,\,\,2NaOH\,\, \to \,\,\,Cu{(OH)_2} \downarrow \,\, + \,\,\,N{a_2}S{O_4} \cr
& (7)\,\,Cu{(OH)_2}\,\xrightarrow{{{t^o}}}\,\,CuO\,\, + \,\,{H_2}O \cr} \)