The brown precipitate in the Nessler test is formed by NH4+ + 2[HgI4]^2- + 4OH- -> HgO.Hg(NH2)I + 7I- + 3H2O. The number of moles of iodide ions liberated per mole of NH4+ is

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The brown precipitate in the Nessler test is formed by $NH X_{4} X^{+} + 2 [HgI X_{4}] X^{2 -} + 4 OH X^{-} HgO \cdot Hg (NH X_{2}) I + 7 I X^{-} + 3 H X_{2} O$ . The number of moles of iodide ions liberated per mole of $NH X_{4} X^{+}$ is

From the balanced equation (Hg 2, I 8, charge

- 7

on both sides), 8 iodine atoms enter as 2

[HgI X_{4}] X^{2 -}

; one stays in the product and 7 are released as

I X^{-}

per

NH X_{4} X^{+}