How many moles of acidified K2Cr2O7 is required to liberate 6 moles of I2 from an aqueous solution of I- ?

Question

How many moles of acidified K2Cr2O7 is required to liberate 6 moles of I2 from an aqueous solution of I- ? (A) 2 (B) 1 (C) 0.25 (D) 0.5

Tardigrade Admin · Accepted Answer

The balanced redox reaction will be: K 2 Cr 2 O 7+6 KI +7 H 2 SO 4 arrow 4 K 2 SO 4+3 I2+ Cr 2( SO 4)3+7 H 2 O 3 moles of I2 requires 1 mole of K 2 Cr 2 O 7 Hence, 6 moles of I2 will require 2 moles of K 2 Cr 2 O 7

Q. How many moles of acidified $K_{2} C r_{2} O_{7}$ is required to liberate $6$ moles of $I_{2}$ from an aqueous solution of $I^{-}$ ?

2

1

0.25

0.5

The balanced redox reaction will be:

$K_{2} C r_{2} O_{7} + 6 K I + 7 H_{2} S O_{4} \to$
$4 K_{2} S O_{4} + 3 I_{2} + C r_{2} (S O_{4})_{3} + 7 H_{2} O$

$3$ moles of $I_{2}$ requires $1$ mole of $K_{2} C r_{2} O_{7}$

Hence, $6$ moles of $I_{2}$ will require $2$ moles of $K_{2} C r_{2} O_{7}$

Q. How many moles of acidified K2​Cr2​O7​ is required to liberate 6 moles of I2​ from an aqueous solution of I− ?

2

1

0.25

0.5

The balanced redox reaction will be: K2​Cr2​O7​+6KI+7H2​SO4​→ 4K2​SO4​+3I2​+Cr2​(SO4​)3​+7H2​O 3 moles of I2​ requires 1 mole of K2​Cr2​O7​ Hence, 6 moles of I2​ will require 2 moles of K2​Cr2​O7​

Q. How many moles of acidified $K_{2} C r_{2} O_{7}$ is required to liberate $6$ moles of $I_{2}$ from an aqueous solution of $I^{-}$ ?

The balanced redox reaction will be:

$K_{2} C r_{2} O_{7} + 6 K I + 7 H_{2} S O_{4} \to$
$4 K_{2} S O_{4} + 3 I_{2} + C r_{2} (S O_{4})_{3} + 7 H_{2} O$

$3$ moles of $I_{2}$ requires $1$ mole of $K_{2} C r_{2} O_{7}$

Hence, $6$ moles of $I_{2}$ will require $2$ moles of $K_{2} C r_{2} O_{7}$