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-

NaI+AgNO_(3)toAgI+NaNO_(2) AgI+Fe to FeI_(2)+Ag FeI_(2)+Cl_(2) to Fe

1.	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-
Answer» 0.5 500 250 0.25 Answer :C