Iodine can be prepared by the following reactions.
`2NaIO_(3) + 5NaSO_(3) rarr 2NaSO_(4) + 2Na_(2) SO_(4) + H_(2)O + I_(2)`
How much `NaIO_(3)` is reuired to produce `127 g` is `I_(2)`?

Options

A. `1.98 kg`
B. `3.96 kg`
C. `5.94 kg`
D. `0.99 kg`

Answer

A

Explanation

`(Mw "of" I_(2) = 245 g, Mw "of" NaIO_(3) - 198 g, Mw "of" NaHSO_(3) = 104 g)`
Mole of `I_(2) = (127)/(254) = 0.5 "mol" I_(2)`
`(0.5 "mol" I_(2)) ((2 "mol" NaIO_(3))/("mol" I_(2)))((198 g "of" NaIO_(3))/("mol of" NaIO_(3)))`
`implies 0.5 xx 2 xx 198 xx 1980 g = 1.98 kg "of" NaIO_(3)`