1 mole of XeF4 on reaction with excess of KI will liberate I2 that requires _________.
A
1 mole of Na2S2O3
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B
2 moles of Na2S2O3
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C
0.5 mole of Na2S2O3
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D
4 moles of Na2S2O3
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Solution
The correct option is B 4 moles of Na2S2O3 XeF4+4KI⟶Xe+2I2+4KF
In the above reaction, two moles of iodine liberates and every mole of iodine requires two moles of hypo solution. So the total solution required is 4 moles.