question_answer

In the equation \[{{H}_{2}}S+2HN{{O}_{3}}\xrightarrow{{}}2{{H}_{2}}O+2N{{O}_{2}}+S\] The equivalent weight of hydrogen sulphide is:

A)  18                                         

B)  68

C)  34                                         

D) (d 17                   

Correct Answer: D

Solution :

                Hence, the equivalent weight of \[{{H}_{2}}S=\frac{Molecular\,weight}{change\,in\,oxidation\,number}\] \[=\frac{34}{2}\] \[=17\]