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2014-07-31T00:03:35+05:30

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Let there be m moles of Fe and n moles of Fe2 O3 initially
Weight W1 = 52 m + (104+48)n = 52 m + 152 n
Let all Fe be converted to Fe2 O3.      4 Fe + 3 O2 -->  2 Fe 2  O3
  So m moles of Fe will interact with  (m/4) * 3 * 32 gm oxygen.  This is the increase in weight
  W2 - W1 = 24 m  = 5% of W1 =>     24 m =  0.05 (52 m + 152 n)
                 21.4 m = 7.6 n  => m : n =  38 : 107
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