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NaI+AgNO(3)toAgI+NaNO(2) AgI+Fe to FeI...

`NaI+AgNO_(3)toAgI+NaNO_(2)`
`AgI+Fe to FeI_(2)+Ag`
`FeI_(2)+Cl_(2) to FeCl_(2)+I_(2)`
(atomic mass of Ag=108, I=127, Fe=56, N=14, Cl=35.5). The above reaction is carried out by taking 75g of NaI and 255 kg of `AgNO_(3)`. Therefore, the number of moles of iodine formed is-

A

0.5

B

500

C

250

D

0.25

Text Solution

Verified by Experts

The correct Answer is:
C
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