An iron ore contain 63 % fe in it find %fe2o3
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atomic mass of iron (fe) = 55.9
atomic mass of oxygen (o)=16
molecular mass of fe2o3 = 55.9×2+16×3
159.8
% of iron in fe203=111.8/159.8×100
= 69.96
= 70% approx
if fe is 70% then % of fe203 = 100
if few is 1% then fe203 = 100/70
if few is 63% then fe2o3% = 100/70×63
=90%
atomic mass of oxygen (o)=16
molecular mass of fe2o3 = 55.9×2+16×3
159.8
% of iron in fe203=111.8/159.8×100
= 69.96
= 70% approx
if fe is 70% then % of fe203 = 100
if few is 1% then fe203 = 100/70
if few is 63% then fe2o3% = 100/70×63
=90%
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