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Volume of Gases and Number of Moles

The amount of...

Question

The amount of $N a_{2} S_{2} O_{3} \cdot 5 H_{2} O$ required to completely reduce $100$ mL of $0.25$ N iodine solution, is:

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Solution

$I_{2} + 2 {N a}_{2} S_{2} O_{3} = 2 N a I + {N a}_{2} S_{4} O_{6}$
$100$ ml $0.25$ N of iodine $(I_{2}) = 25$ millimoles.
$∴$ $25 \times 2 = 50$ millimoles of ${N a}_{2} S_{2} O_{3}$ will be oxidized.
$∴$ $50 \times 248.11$ mg of ${N a}_{2} S_{2} O_{3} = 12.405$ gm of ${N a}_{2} S_{2} O_{3} .5 H_{2} O$ will be oxidized.

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