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Question

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 C 103 kg
The balanced chemcial equations for the preparation of NaBr are as follows:
Fe+Br2FeBr2 .....(i)
3FeBr2+Br2Fe3Br8 .....(ii)
Fe3Br8+4Na2CO38NaBr+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

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