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) \[N{{a}_{2}}{{S}_{2}}{{O}_{3}},C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]
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\[S{{O}_{3}}\]and\[S{{O}_{2}}\]is obtained along with yellow precipitate when thiosulphate 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 \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}+2HCl\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} suffocating\text{ }gas \\ \,\,\,\,\,\,\,\,\,\,\,(B) \end{smallmatrix}}{\mathop{S{{O}_{2}}}}\,\]\[+2NaCl+{{H}_{2}}O+~\underset{yellow}{\mathop{S}}\,\] \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+3S{{O}_{2}}+{{H}_{2}}S{{O}_{4}}\xrightarrow[{}]{{}}{{K}_{2}}S{{O}_{4}}\]\[+\underset{green}{\mathop{C{{r}_{2}}{{(S{{O}_{4}})}_{3}}}}\,+{{H}_{2}}O\]
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