$S_{2} O_{8}^{2 -} + I^{-} S O_{4}^{2 -} + I_{2}$
What is the maximum yield of iodine that can be obtained when 1 mole of $N a_{2} S_{2} O_{8}$ reacts completely with excess iodide ion according to the equation above? (Equation above is not balanced.)

1 moles

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2 moles

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4 mole

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6 moles

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8 moles

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Solution

The correct option is A 1 moles
The unbalanced chemical equation is,
$. . . S_{2} O_{8}^{2 -} + . . . I^{-} \to . . . S O_{4}^{2 -} + . . . I_{2}$
The balanced chemical equation is
$S_{2} O_{8}^{2 -} + 2 I^{-} \to 2 S O_{4}^{2 -} + I_{2}$
$1$ mole of $S_{2} O_{8}^{2 -}$ gives $1$ mole of $I_{2}$
The maximum yield of iodine that can be obtained when $1$ mole of $S_{2} O_{8}^{2 -}$ reacts completely with excess iodide ion according to the equation above is $1$ mole.

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