Question

Fe203 contain 63%. Find iron content

Answer 1

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%

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Answer 2

Answer:

atomic mass of iron (fe) = 55.9

atomic mass of oxygen (o)=16

molecular mass of fe203 = 55.9x2+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 fe203% = 100/70×63

=90%