Question

A mixture of FeO and Fe304 when heated in air to constant weight gains 5% in its weight. The % of FeO in the sample

Joshi sir comment

let FeO = x gm and FeO+Fe2O3= y gm so FeO = [72/(72+160)]*y and Fe2O= [160/(160+72)]*y

2FeO + 1/2O2  ---------->  Fe2O3

so amount of Fe2Oformed by x gm FeO  = 160x/144

similarly amount of Fe2Oformed by  [72/(72+160)]*y gm FeO = 160 [72/(72+160)]*y/144

so total Fe2O3 = 160x/144 + 160 [72/(72+160)]*y/144 +  [160/(160+72)]*y = 160x/144 + [160*72/232]y/144 + 160y/232

now according to the given condition {[160x/144 + [160*72/232]y/144 + 160y/232}/{x+y) = 105/100

now solve

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