A) \[N{{a}_{2}}S{{O}_{3}},\,S{{O}_{2}}\,\,C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]
B) \[N{{a}_{2}}{{S}_{2}}{{O}_{3}},\,S{{O}_{2}}\,\,C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]
C) \[N{{a}_{2}}S,\,\,S{{O}_{2}},\,\,\,C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]
D) \[N{{a}_{2}}S{{O}_{4}},\,\,S{{O}_{2}},\,C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]
Correct Answer: B
Solution :
Gas B turns the colour of acidified \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\]green thus it is \[S{{O}_{2}}\] and \[S{{O}_{2}}\]is obtained along with yellow precipitate when thio sulphate is treated with dilute acids. Thus, A is \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\], B is \[S{{O}_{2}}\] and C is \[C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]. The reactions are as follows \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}+HCl\xrightarrow{{}}\underset{\begin{smallmatrix} suffocating \\ gas \\ (B) \end{smallmatrix}}{\mathop{S{{O}_{2}}}}\,\] \[+\,NaCl+{{H}_{2}}O+\underset{yellow}{\mathop{S}}\,\] \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+S{{O}_{2}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}\] \[+\,\,\underset{green}{\mathop{C{{r}_{2}}{{(S{{O}_{4}})}_{3}}}}\,+{{H}_{2}}O\]
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