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`
A
`420 g`
B
`420 kg`
C
`4.2 xx 10^(5) kg`
D
`4.2 xx 10^(8) g`
Text Solution
Verified by Experts
The correct Answer is:
B
`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`
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