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2Na(2)S(2)O(3)+I(2)toNa(2)S(4)O(6)+2NaI ...

`2Na_(2)S_(2)O_(3)+I_(2)toNa_(2)S_(4)O_(6)+2NaI`
How many equivalents of Hypo is oxidised by one mole of Iodine?

Text Solution

Verified by Experts

The correct Answer is:
2
`2Na_(2)overset(+2)(S_(2))O_(3)+overset(0)I_(2)rarrNa_(2)overset(+2.5)(S_(4))O_(6)+2NaI^(-1)`
EQ wt of `Na_(2)S_(2)O_(3)=("Mol wt of"Na_(2)S_(2)O_(3))/(1)`
EQ wt of Iodine = `("Mol wt of iodine")/(2)`
1 mole iodine = 2 equivalents iodine
= 2 equivalents hypo
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