Question

$4I^{\ominus }+H{g}^{2+}\to Hg{I}_4^{2-}$
$1$ mol each of $H{g}^{2+}$ and $I^{\ominus }$ will form $Hg{I}_4{}^{2-}$

• A. $1mol$
• B. $0.5mol$
• C. $0.25mol$
• D. $2mol$

Answer 1

Answer: (C)

$I^{\ominus }$ is limiting reagent, so one mole $I^{\ominus }$ will give $\frac{1}{4}$ mol or $0.25$ mole of $Hg{I}_4{}^{2-}$