`4I^(-)+Hg^(2+)rarrHgI_(4)^(2-)` , `1` mole each of `Hg^(2+)` and `I^(-)` will form….. Mole `HgI_(4)^(2-)`:A. 1 molB. 0.5 molC. 0.25 molD. 2 mol

`4I^(-)+Hg^(2+)rarrHgI_(4)^(2-)` , `1` mole each of `Hg^(2+)` and `I^(

1.	`4I^(-)+Hg^(2+)rarrHgI_(4)^(2-)` , `1` mole each of `Hg^(2+)` and `I^(-)` will form….. Mole `HgI_(4)^(2-)`:A. 1 molB. 0.5 molC. 0.25 molD. 2 mol
Answer» Correct Answer - C `I^(ɵ)` is limiting reagent so one mole `I^(ɵ)` will give `(1)/(4)` mol or `0.25` moles of `Hgl_(4)^(2-)`

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`4I^(-)+Hg^(2+)rarrHgI_(4)^(2-)` , `1` mole each of `Hg^(2+)` and `I^(-)` will form….. Mole `HgI_(4)^(2-)`:A. 1 molB. 0.5 molC. 0.25 molD. 2 mol

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