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Comprehension # 3 NaBr, used to produ...

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)`
If yield of `(iii)` reaction is `90%` then mole of `CO_(2)` formed when `2.06xx10^(3) kg NaBr` is formed.

A

`20`

B

`10`

C

`40`

D

none

Text Solution

Verified by Experts

The correct Answer is:
B
mole of `CO_(2)=("mole of NaBr")/(2)`
`=(2.06xx10^(3))/(103xx2)=10`
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