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Anomalous Behavior of Lithium Halogens

2NaIO3+5NaHSO...

Question

$2 N a I O_{3} + 5 N a H S O_{3} \to 2 N a H S O_{4} + 2 N a_{2} S O_{4} + H_{2} O + I_{2}$

if the given reaction is used for the preparation of iodine gas then how much $N a H S O_{3}$ is required to produce $381$ g of $I_{2}$ ?

156.0

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390.0

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520.0

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780.0

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Solution

The correct option is D $780.0$ g
The molar masse of $N a H S O_{3}$ is 104 g/mol and iodine is 253.8 g/mol respectively.

From the balanced chemical equation, it can be seen that 5 moles of $N a H S O_{3}$ give 1 mole of iodine.
Hence, $5 \times 104 = 520$ g of $N a H S O_{3}$ will give 253.8 g of iodine.
Hence, 381 g of iodine will be prepared by $\frac{520}{253.8} \times 381 = 780$ g of $N a H S O_{3}$

Hence, the option $(D)$ is correct.

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