`4I^(-)+Hg^(2+)rarrHgI_(4)^(2-)` , `1` mole each of `Hg^(2+)` and `I^(-)` will form….. Mole `HgI_(4)^(2-)`:A. `1 "mole"`B. `0.5 "mole"`C. `0.25 "mole"`D. `2 "mole"`

`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 "mole"`B. `0.5 "mole"`C. `0.25 "mole"`D. `2 "mole"`
Answer» Correct Answer - C `4` mole of `I^(-)` will give `1` mole of `HgI_(4)^(2-)` with `1` mole of `Hg^(2+)`. The mole ratio order is `4:1::1`

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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 "mole"`B. `0.5 "mole"`C. `0.25 "mole"`D. `2 "mole"`

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