A) 18
B) 68
C) 34
D) 17
Correct Answer: D
Solution :
\[{{H}_{2}}S+2HN{{O}_{3}}\xrightarrow{{}}2{{H}_{2}}O+2N{{O}_{2}}+S\] Hence, the equivalent weight of \[{{H}_{2}}S=\frac{\text{mlecular}\,\text{weight}}{\text{change}\,\text{in}\,\text{oxidation}\,\text{number}}=\frac{34}{2}=17\]You need to login to perform this action.
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