You visited us 1 times! Enjoying our articles? Unlock Full Access!

Limiting Reagent or Reactant

Which of the ...

Question

Which of the following reactions is/are correct?

C_{x} H_{y} + (x + \frac{y}{2}) O_{2} ⟶ x C O_{2} + \frac{y}{2} H_{2} O

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

4 F e^{3 +} + [F e (C N)_{6}]^{4 -} ⟶ F e_{3} [F e (C N)_{6}]_{4}

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

5 C O + I_{2} O_{5} ⟶ I_{2} + 5 C O_{2}

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

P b^{2 +} + S^{2 +} ⟶ P b S

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

Open in App

Solution

The correct option is C $5 C O + I_{2} O_{5} ⟶ I_{2} + 5 C O_{2}$
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} ⟶ x C O_{2} + \frac{y}{2} H_{2} O$

$4 F e^{3 +} + 3 [F e (C N)_{6}]^{4 -} ⟶ F e_{4} [F e (C N)_{6}]_{3}$

$5 C O + I_{2} O_{5} ⟶ I_{2} + 5 C O_{2}$

$P b^{2 +} + S^{2 -} ⟶ P b S$

Hence, option C is correct.

Suggest Corrections

Similar questions

[F e (C N)_{6}]^{3 -}

[F e (C N)_{6}]^{4 -}

E^{\circ}

[F e (C N)_{4}]^{4^{-}} \to [F e (C N)_{6}]^{3 -} + e^{-}

\begin{matrix} E^{\circ} = - 0.35 V F e^{2 +} \to F e^{3 +} + e^{-} & E^{\circ} = - 0.77 V \end{matrix}

K_{4} [F e {(C N)}_{6}] O x i d a t i o n - ------------- \to {F e}^{3 +} + C O_{2} + N O_{3}^{⊖}

C O

I_{2} O_{5} + 5 C O \to I_{2} + 5 C O_{2} .

127.0 g

I_{2}

70 g

C O

C O

C u C l

E^{o}

{[F e {(C N)}_{6}]}^{4 -} ⟶ {[F e {(C N)}_{6}]}^{3 +} + e^{-}, E^{o} = - 0.35 V

{F e}^{2 +} \to {F e}^{3 +} + e^{- 1}

{F e}^{2 +} \to {F e}^{3 +} + e^{- 1}; E^{o} = - 0.77 V

Join BYJU'S Learning Program