When CrI 3 is treated with Cl 2 in basic medium- the following reaction takes place- CrI 3- KOH - Cl 2- K 2 Cr 2 O 7- KIO 4- KCl - H 2 OTo produce 81 mols of potassium chloride- how many mols of CrI 3 should be taken-

Balanced reaction is given: 2CrI_3 + 64KOH + 27 Cl_2 → 2K_2 Cr_2 O_7 + 6KIO_4 + 54KCl + 32 H_2 O To produce 54 moles of KCl we need 2 moles of CrI_3 So, for 81 ...

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Standard XII

Chemistry

Types of Redox Reactions

When CrI 3 is...

Question

When $C r I_{3}$ is treated with $C l_{2}$ in basic medium, the following reaction takes place.
$C r I_{3} + K O H + C l_{2} \to K_{2} C r_{2} O_{7} + K I O_{4} + K C l + H_{2} O$
To produce 81 mols of potassium chloride, how many mols of $C r I_{3}$ should be taken?

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Solution

Balanced reaction is given:
$2 C r I_{3} + 64 K O H + 27 C l_{2} \to 2 K_{2} C r_{2} O_{7} + 6 K I O_{4} + 54 K C l + 32 H_{2} O$
To produce 54 moles of $K C l$ we need 2 moles of $C r I_{3}$
So, for 81 moles we will need $\frac{81 \times 2}{54} = 3$

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When CrI3 is treated with Cl2 in basic medium, the following reaction takes place. CrI3+KOH+Cl2→K2Cr2O7+KIO4+KCl+H2O To produce 81 mols of potassium chloride, how many mols of CrI3 should be taken?

Balanced reaction is given: 2CrI3+64KOH+27Cl2→2K2Cr2O7+6KIO4+54KCl+32H2O To produce 54 moles of KCl we need 2 moles of CrI3 So, for 81 moles we will need 81×254=3

When $C r I_{3}$ is treated with $C l_{2}$ in basic medium, the following reaction takes place.
$C r I_{3} + K O H + C l_{2} \to K_{2} C r_{2} O_{7} + K I O_{4} + K C l + H_{2} O$
To produce 81 mols of potassium chloride, how many mols of $C r I_{3}$ should be taken?

Balanced reaction is given:
$2 C r I_{3} + 64 K O H + 27 C l_{2} \to 2 K_{2} C r_{2} O_{7} + 6 K I O_{4} + 54 K C l + 32 H_{2} O$
To produce 54 moles of $K C l$ we need 2 moles of $C r I_{3}$
So, for 81 moles we will need $\frac{81 \times 2}{54} = 3$