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What is the equivalent weight of `NH_(3)` in the given reaction? `3CuO+2NH_(3) rarr 3Cu+N_(2)+3H_(2)O`A. `17`B. `(17)/(4)`C. `(17)/(2)`D. `(17)/(3)` |
Answer» Correct Answer - D `2N^(3-) rarr (N_(2))^(0)+6e^(-)` `2` mole of `NH_(3)=1` mole `N_(2)` Thus equivalents `2xxn=1xx6` `n=6//2=3` `:.` Eq. wt. `= (M)/(3)=(17)/(3)` |
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