A MIXTURE OF Fe AND Fe2O3 when heated in air gains 5% in weight . Find the composition of the initial mixture?
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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
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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