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 \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}+HCl\xrightarrow{{}}\underset{\begin{smallmatrix} suffocating\,gas \\ (B) \end{smallmatrix}}{\mathop{S{{O}_{2}}}}\,\] \[+NaCl+{{H}_{2}}O+\underset{yellow}{\mathop{S}}\,\] \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+S{{O}_{2}}+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}\] \[+C{{r}_{2}}\underset{green}{\mathop{{{(S{{O}_{4}})}_{3}}+{{H}_{2}}O}}\,\]
You need to login to perform this action.
You will be redirected in
3 sec
