Question

Question #57ca8

Answer 1

The mass of the nail before corrosion is 0.25g.

The volumes and concentration of the acid are irrelevant as they are in excess.

The iron in the nail has been dissolved by the acid to form `Fe_((aq))^(2+)`

We first use the titration result to find the number of moles of iron and from that, the mass.

`c=(n)/(v)`

Where `c` = concentration
`n` = number of moles
`v` = volume in `dm^3` (that's the same as litre or liter)

Start with the equation:

`MnO_(4)^(-)+8H^(+)+5Fe^(2+)rarrMn^(2+)+4H_2O+5Fe^(3+)`

This tells us that for every mole of `MnO_4^(-)` there must be 5 moles of `Fe^(2+)`

Since `c=(n)/(v)` then `n=cv`

So the number of moles of `MnO_4^(-)` = 0.02 x 0.025 = 5 x `10^(-4)` (note that I had to convert 25 `cm^3` to 0.025 `dm^3` by dividing by 1000)

From the equation the number of moles of `Fe^(2+)` = 5 x `10^(-4)` x 5 = 2.5 x `10^(-3)`

The `A_r` of Fe = 56 so the mass of iron = 2.5 x `10^(-3)` x 56 = 0.14g

The total mass of the rusty nail = 0.3 g so the mass of the rust = 0.3 - 0.14 = 0.16g

Rust is hydrated iron (III) oxide `Fe_3O_3.nH_2O` The question does not give you its formula so I'll just assume it to `Fe_2O_3`

The `M_r` of `Fe_2O_3` = (2 x 56) + (3 x 16) = 160

So the number of moles of `Fe_2O_3` = 0.16/160 = 0.001

So the number of moles of Fe present must be 0.001 x 2 =0.002

So the mass of Fe present must be 0.002 x 56 = 0.112g

So the total mass of Fe in the shiny new nail must be 0.14 + 0.112 = 0.252 which I'll round to 0.25g

This is a badly worded question for 2 reasons:

1. The concentration and volume of the acid are not needed in the calculation which may confuse students. It would be better to say "excess acid"

2. The correct formula for rust should have been given.