In the following sequence of reactions:
$C a O C l_{2} + H_{2} O ⟶ C a (O H)_{2} + C l_{2}$
$2 K I + C l_{2} ⟶ 2 K C l + I_{2}$
$I_{2} + 2 N a_{2} S_{2} O_{3} ⟶ N a_{2} S_{4} O_{6} + 2 N a I$
Calculate moles of $N a_{2} S_{2} O_{3}$ required for complete reduction of $I_{2}$ obtained from $127$ g $C a O C l_{2}$ sample ( $50 %$ purity).

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Solution

The molar mass of $C a O C l_{2}$ is $127.1$ g/mol.
Thus, $127$ g of $50 %$ pure $C a O C l_{2}$ corresponds to $0.5$ moles.
They will produce $0.5$ moles of chlorine which in turn will produce $0.5$ moles of iodine.
$0.5$ moles of iodine will be reduced by $1$ mole of $N a_{2} S_{2} O_{3}$ .

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M n O_{2} + 4 H C l \to M n C l_{2} + C l_{2} + 2 H_{2} O

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C a O C l_{2} + H_{2} O \to C a (O H)_{2} + C l_{2}

C r O_{2} C l_{2} + 2 N a O H \to N a_{2} C r O_{4} + 2 H C l

C a (O C l)_{2} + 2 C H_{3} C O O H \to C a (C H_{3} C O O)_{2} + H_{2} O + C l_{2}

C l_{2} + 2 K I \to 2 K C l + I_{2}

2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I

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N a_{2} S_{2} O_{3}

2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I

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