The following process has been used to obtain iodine from oil-field drines in California.
`NaI+AgNO_(3) rarr AgI+NaNO_(3)` …(1)
`2AgI+Fe rarr FeI_(2)+2Ag …(2)`
`2FeI_(2)+3Cl_(2) rarr 2FeCl_(3)+2I_(2)` …(3)
How many grams of `AgNO_(3)` are required in the first step for every `254 kg I_(2)` produced in the third step?

Options

A. (a)`340xx10^(3)`
B. (b)`240xx10^(3)`
C. (c )`440xx10^(3)`
D. (d)`540xx10^(3)`

Answer

A

Explanation

Balanced equation:
`Nal+AgNO_(3)rarr Agl+NaNO_(3) …(1)`
`AgI+Fe rarr FeI_(2)+2Ag …(2)`
`2FeI_(2)+3Cl_(2) rarr 2FeCl_(2)+2I_(2) …(3)`
From eq. (3), we get
`("mole of" I_(2))/2=("mole of" FeI_(2))/2`
`("mole of" FeI_(2))/1=("mole of" AgI)/2`
`("mole of" AgI)/1=("mole of" AgNO_(3))/1`
`:.` mole of `I_(2)=("mole of" FeI_(2))`
`(("mole of" AgI)/2)=(("mole of" AgNO_(3))/2)`
`(254xx10^(3))/254=("mole of" AgNO_(3))/2`
`2xx10^(3)`= mole of `AgNO_(3)`
`=("mass of" AgNO_(3))/("molar mass of" AgNO_(3))`
Mass of `AgNO_(3)=170xx(2xx10^(3))g=340xx10^(3) g`