Comprehension # 3
`NaBr`, used to produce `AgBr` for use in photography can be self prepared as follows :
`{:(,Fe+Br_(2)rarrFeBr_(2)," "..........(i),),(,FeBr_(2)+Br_(2)rarrFe_(3)Br_(8)," " ..........(ii) ," "("not balanced")),(,Fe_(3)Br_(8)+Na_(2)CO_(3)rarrNaBr+CO_(2)+Fe_(3)O_(4) , " ".......(iii)," "("not balanced")):}`
How much `Fe` in `kg` is consumed to produce `2.06xx10^(3) kg Nabr " "........(iv)`
Mass of iron required to produce `2.06xx10^(3) kg NaBr`

Options

A. `420 g`
B. `420 kg`
C. `4.2 xx 10^(5) kg`
D. `4.2 xx 10^(8) g`

Answer

B

Explanation

`8` mole `NaBr` obtain from `=3` mole `Fe`
mole of `Fe` = mole `NaBr = (2.06xx10^(3))/(103xx8)xx3
mass of `Fe` = (2.06xx10^(3))/(103)xx56xx(3)/(8) = 420 kg`