The equivalent weight of `HNO_(3)` (molecular weight `=63`) in the following reaction is `3Cu+8HNO_(3)rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O`A. `(4xx63)/(3)`B. `(63)/(5)`C. `(63)/(3)`D. `(63)/(8)`

The equivalent weight of `HNO_(3)` (molecular weight `=63`) in the fol

1.	The equivalent weight of `HNO_(3)` (molecular weight `=63`) in the following reaction is `3Cu+8HNO_(3)rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O`A. `(4xx63)/(3)`B. `(63)/(5)`C. `(63)/(3)`D. `(63)/(8)`
Answer» Correct Answer - D In this case only `2 mol` of `NO_(3)^(ө)` undergo reduction. `3e^(-)+NO_(3)^(ө)rarrNO(x=3)` (reduction) `x-6= -1` `x-2=0` `x=5, x=2` `6 mol` of `HNO_(3)` are not changing so `6NO_(3)^(ө)` are added in the reaction to get `3 mol` of `Cu(NO_(3))_(2)`. `:. Ew=M+(M)/(3)=(4M)/(3)=(4xx63)/(3)`

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The equivalent weight of `HNO_(3)` (molecular weight `=63`) in the following reaction is `3Cu+8HNO_(3)rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O`A. `(4xx63)/(3)`B. `(63)/(5)`C. `(63)/(3)`D. `(63)/(8)`

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