I.某无色透明溶液可能含有Cu2+、Ca2+、K+、Ag+、SO
、CO
、Cl-中的几种,现进行如下实验:
①滴加BaCl2溶液,有白色沉淀产生,将沉淀滤出。
②向上述沉淀中加入足量的稀硝酸,沉淀部分溶解。
③向①滤液加入AgNO3溶液,有白色沉淀生成,该沉淀不溶于稀HNO3。
(1)试判断:该溶液中肯定有{#blank#}1{#/blank#},肯定没有{#blank#}2{#/blank#},可能有{#blank#}3{#/blank#}(填离子符号)。
(2)写出步骤②发生反应的离子方程式{#blank#}4{#/blank#}。
(3)若要检验溶液中可能存在的离子是否存在,可以进行的实验操作为{#blank#}5{#/blank#}。
Ⅱ.完成下列问题
(4)某溶液中含有较大量的Cl
-、CO
![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsubsup%3E%3Cmrow%3E%3C%2Fmrow%3E%3Cmtext%3E3%3C%2Fmtext%3E%3Cmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3Cmo%3E%E2%88%92%3C%2Fmo%3E%3C%2Fmrow%3E%3C%2Fmsubsup%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
、OH
- , 如果只取一次该溶液就能够分别将3种阴离子依次检验出来,下列实验操作顺序正确的是{#blank#}6{#/blank#}。(填写序号,可重复使用)
①滴加Mg(NO3)2溶液;②过滤;③滴加AgNO3溶液;④滴加Ba(NO3)2溶液
Ⅲ.下面三个方法都可以用来制氯气:
①![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmtext%3E4HCl%28%3C%2Fmtext%3E%3Cmi+mathvariant%3D%22normal%22%3E%E6%B5%93%3C%2Fmi%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3E%29%2BMnO%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmtable%3E%3Cmtr%3E%3Cmtd%3E%3Cmrow%3E%3Cmunder+accentunder%3D%22false%22%3E%3Cmunder+accentunder%3D%22false%22%3E%3Cmtext%3E%CE%94%3C%2Fmtext%3E%3Cmo+stretchy%3D%22true%22%3E_%3C%2Fmo%3E%3C%2Fmunder%3E%3Cmo+stretchy%3D%22true%22%3E_%3C%2Fmo%3E%3C%2Fmunder%3E%3C%2Fmrow%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3Cmtr%3E%3Cmtd%3E%3Cmrow%3E%3C%2Fmrow%3E%3C%2Fmtd%3E%3C%2Fmtr%3E%3C%2Fmtable%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3EMnCl%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3E%2BCl%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmo%3E%E2%86%91%3C%2Fmo%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3E%2B2H%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmtext%3EO%3C%2Fmtext%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
②![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3EKClO%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E3%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmtext%3E%2B6HCl%28%3C%2Fmtext%3E%3Cmi+mathvariant%3D%22normal%22%3E%E6%B5%93%3C%2Fmi%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3E%29%3D3C1%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmo%3E%E2%86%91%3C%2Fmo%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3E%2BKCl%2B3H%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmtext%3EO%3C%2Fmtext%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
③![](http://math.21cnjy.com/mml2svg?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3EKMnO%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E4%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmtext%3E%2BHCl%28%3C%2Fmtext%3E%3Cmi+mathvariant%3D%22normal%22%3E%E6%B5%93%3C%2Fmi%3E%3Cmtext%3E%29%3C%2Fmtext%3E%3Cmo%3E%E2%86%92%3C%2Fmo%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3EKCl%2BMnCl%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3E%2BCl%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmo%3E%E2%86%91%3C%2Fmo%3E%3Cmsub%3E%3Cmrow%3E%3Cmtext%3E%2BH%3C%2Fmtext%3E%3C%2Fmrow%3E%3Cmtext%3E2%3C%2Fmtext%3E%3C%2Fmsub%3E%3Cmtext%3EO%28%3C%2Fmtext%3E%3Cmi+mathvariant%3D%22normal%22%3E%E6%9C%AA%3C%2Fmi%3E%3Cmi+mathvariant%3D%22normal%22%3E%E9%85%8D%3C%2Fmi%3E%3Cmi+mathvariant%3D%22normal%22%3E%E5%B9%B3%3C%2Fmi%3E%3Cmtext%3E%29%3C%2Fmtext%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
根据以上三个反应,回答下列有关问题:
(5)反应②中,氧化产物与还原产物的质量比为{#blank#}7{#/blank#}。
(6)若要制得相同质量的氯气,①②③反应中电子转移的数目之比为{#blank#}8{#/blank#}。