Question

Given a reaction: $4I^{⊝}+H{g}^{2+}\to Hg{I}_4^{2-}$
$1$ mol each of $H{g}^{2+}$ and $I^{⊝}$ will form $Hg{I}_4^{2-}$ equal to:

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

Answer 1

Answer: (C)

$0.25$ mol

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