If the yield of reaction (ii) is 60% and reaction (iii) is 70%, then mass of iron required to produce 2.06×103 kg NaBr is :
A
106 kg
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B
105 kg
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C
103 kg
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D
none of the above
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Solution
The correct option is C103 kg The balanced chemcial equations for the preparation of NaBr are as follows: Fe+Br2⟶FeBr2 .....(i) 3FeBr2+Br2⟶Fe3Br8 .....(ii) Fe3Br8+4Na2CO3⟶8NaBr+4CO2+Fe3O4 .....(iii) From the balanced chemical equation, it can be seen that 8 moles of NaBr are produced from 3 moles of Fe. The molar masses of Fe and NaBr are 55.8 g/mol and 103 g/mol respectively. Thus, mass of Fe in kg is consumed to produce 2.06×103 kg NaBr
=2.06×103 kg NaBr×3×55.88×103=420 kg Fe
If the yield of reaction (ii) is 60% and reaction (iii) is 70%. then mass of iron required to produce 2.06×103 kg NaBr will be 420×10060×10070=103 kg