Question

If a piece of iron gains 10% of its weight due to partial rusting into Fe2O3? then the percentage of total iron that has rustedis

Answer 1

initial mass of iron = x
mole of iron converted to rust = y
mass of iron not got rusted = x - 56y
mass of y converted to rust= moles x molar mass 
= 56y

4 moles of iron form 2 moles of rust
moles of rust formed = y/2
mass of rust formed = y/2 x 160 = 80y

mass of iron not converted to rust and mass of rust has a total mass of x & 10% x
or x + 0.1x = 1.1x

the equation will be 
x - 56y = 1.1x
24y = 0.1x
y = 0.1x/24

by putting the value of y in equatio 56y we get
= 56 x 0.1x/24

% of iron converted to rust = mass of rust / initial mass x 100
= 56 x 0.1x/ 24 x x x 100
= 23.3%

Answer 2

Answer:

4Fe + 3O2----2Fe2O3

let the weight of fe converted to rust be x gm

initial total weight be 100gm

now weight gain=110gm

mole of fe converted to fe2o3

=4x/56

mole of fe2o3=x/56*2

weight of fe2o3=10x/7

now total weight is 110 gm

therefore

100-x +10x/7=110gm

x=23.3

Hope it helps....