A) \[N{{a}_{2}}C{{O}_{3}}\] and \[NaCl\]
B) \[N{{a}_{2}}S{{O}_{3}}\] and \[N{{a}_{2}}S\]
C) \[N{{a}_{2}}S\] and \[N{{a}_{2}}S{{O}_{3}}\]
D) \[N{{a}_{2}}S{{O}_{3}}\] and \[N{{a}_{2}}S{{O}_{4}}\]
Correct Answer: B
Solution :
Gas (Y) is \[S{{O}_{2}}\] obtained from \[N{{a}_{2}}S{{O}_{3}}(A)\] and gas (Z) is \[{{H}_{2}}S\] obtained from \[N{{a}_{2}}S(B)\]. \[\underset{(A)}{\mathop{N{{a}_{2}}S{{O}_{3}}}}\,+2HCl\xrightarrow{{}}2NaCl+\underset{(Y)}{\mathop{S{{O}_{2}}}}\,+{{H}_{2}}O\] \[\underset{(B)}{\mathop{N{{a}_{2}}S}}\,+2HCl\xrightarrow{{}}2NaCl+\underset{(Z)}{\mathop{{{H}_{2}}O}}\,\] \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+{{H}_{2}}S{{O}_{4}}+3S{{O}_{2}}\xrightarrow{{}}{{K}_{2}}S{{O}_{4}}\] \[+\underset{green}{\mathop{C{{r}_{2}}{{(S{{O}_{4}})}_{3}}}}\,+{{H}_{2}}O\] \[Pb{{(C{{H}_{3}}COO)}_{2}}+{{H}_{2}}S\xrightarrow{{}}\underset{black}{\mathop{PbS}}\,\downarrow +2C{{H}_{3}}COOH\]
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