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4I^(-)+Hg^(2+)rarrHgI(4)^(2-) , 1 mole e...

`4I^(-)+Hg^(2+)rarrHgI_(4)^(2-)` , `1` mole each of `Hg^(2+)` and `I^(-)` will form….. Mole `HgI_(4)^(2-)`:

A

1 mol

B

0.5 mol

C

0.25 mol

D

2 mol

Text Solution

Verified by Experts

The correct Answer is:
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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HgCl_(2)+4KI to K_(2)[H_gI_(4)]+2KCl 1 mole each of Hg^(2+) and I^(-) will form how many moles of HgI_(4)^(2-)?

HgI_(2)darr+2HI to K_(2)[HgI_(4)]

HgI_(2)darr+KI hArrK_(2)[HgI_(4)]

The solubility product of Hg_(2)I_(2) is equal to

How many moles of HgI_4^(2-) will be formed when 2 " mol of " Hg^(2+) and 2 mol I^(ɵ) react according to the following equation? Hg^(2+)+4I^(ɵ)toHgI^(2-) (a). 1 mol (b). 0.5 mol (c). 0.25 mol (d). 2 mol

2KI+HgI_(2) darr to K_(2)[HgI_(4)]

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