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Question

NaBr, used to produce AgBr for use in photography can itself be prepared as follows:
Fe+Br2FeBr2
FeBr2+Br2Fe3Br8 (not balanced)
Fe3Br8+Na2CO3NaBr+CO2+Fe3O4 (not balanced)
How much Fe, in kg, is consumed to produce 4.12×103 kg NaBr

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Solution

On converting the given amount of NaBr into mole =WM=4.12×103×103g(23+80)=4.12×106103mol
Balanced reaction
(i)(Fe+Br2FeBr2)×3
3Fe+3Br23FeBr2
(ii) 3FeBr2+Br2Fe3Br8
(iii) Fe3Br8+4Na2CO38NaBr+4CO2+Fe3O4
From stoichiometry of all the reaction 8 mole of NaBr is obtained from 1 mole of Fe3Br8. Which is obtained from 3 mole of FeBr2 in previous reaction.
According to first reaction 3 mole of FeBr2 is obtained from 3 mole of Fe. Hence, we can say that 8 mole of NaBr will be obtained from 3 mole of Fe.
8 mole of NaBr is obtained from = 3 mole of Fe
4.12×106103 mole of NaBr will be obtained from =38×4.12×106103 mole of Fe
wt. of required Fe=n×molar mass
38×4.12×106103×56=840 kg

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