If a piece of iron gains 10% of its mass due to partial rusting into Fe2O3 the percentage of total iron that has rusted is

Asked by Anil | 11th Jun, 2017, 07:56: PM

Expert Answer:

The chemical reaction is as follows: 
4Fe + 3O2 → 2Fe2O3
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 equation 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 100
= 23.3%

Answered by Prachi Sawant | 15th Jun, 2017, 05:07: PM

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