$H g C l_{2} + 4 K I \to K_{2} [H g I_{4}] + 2 K C l$
1 mole each of $H g^{2 +}$ and $I^{-}$ will form how many moles of $H g I_{4}^{2 -}$ ?

1 mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5 Mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.25 mol

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 mol

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 0.25 mol
ANS; C
here in given condition limiting agent is KI.
according to reaction mole of $H g I_{4}^{2 -}$ = $\frac{1}{4}$ of mole of fKI
mole of $H g I_{4}^{2 -}$ = $\frac{1 \times 1}{4}$ = 0.25 mole

Suggest Corrections

Similar questions

4 I^{⊝} + H g^{2 +} \to H g I_{4}^{2 -}

1

H g^{2 +}

I^{⊝}

H g I_{4}^{2 -}

M g_{3} (P O_{4})_{2}

M n O_{4}^{⊝}

1

H g C l_{2}

I_{2}

I^{-}

Join BYJU'S Learning Program