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Question


2NaIO3+5NaHSO32NaHSO4+2Na2SO4+H2O+I2

if the given reaction is used for the preparation of iodine gas then how much NaHSO3 is required to produce 381 g of I2?


A
156.0 g
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B
390.0 g
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C
520.0 g
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D
780.0 g
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Solution

The correct option is D 780.0 g
The molar masse of NaHSO3 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 NaHSO3 give 1 mole of iodine.
Hence, 5×104=520 g of NaHSO3 will give 253.8 g of iodine.
Hence, 381 g of iodine will be prepared by 520253.8×381=780 g of NaHSO3

Hence, the option (D) is correct.

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