Question

calculate the amount of rust (Fe2O3.3H2O) formed by complete rusting of 100 kg of iron

Answer 1

Answer:

The atomic weights we need are those of iron=55.845

For oxygen=16.00

For hydrogen=1.008

For every 2*55.845=111.69grams of iron

The mass of Fe2O3.3H2O will be

111.69+3*16.00+3*(2.016+16.00)=213.74grams

If when 100 kg of iron has turned to rust the mass of the type of rust described here is (100kg)*(213.74/111.69)=191.36kg