Question

A student determines the iron(III) content of a solution by first precipitating it as iron(III) hydroxide, and then decomposing the hydroxide to iron(III) oxide by heating. How many grams of iron(III) oxide should the student obtain if her solution contains 52.0 mL of 0.404 M iron(III) nitrate?

Answer 1

3.3535 g

Step-by-step explanation:

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

Or,

`Moles =Molarity * {Volume\ of\ the\ solution}`

Given :

For iron(III) nitrate :

Molarity = 0.404 M

Volume = 52.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 52.0×10⁻³ L

Thus, moles of iron(III) nitrate :

`Moles=0.404 * {52.0* 10^(-3)}\ moles`

Moles of iron(III) nitrate = 0.021 moles

1 mole of iron(III) nitrate forms 1 mole of iron(III) hydroxide which further forms 1 mole of iron(III) oxide.

Thus, moles of iron(III) oxide formed from 0.021 moles of iron(III) nitrate = 0.021 moles

Also, molar mass of iron(III) oxide = 159.69 g/mol

So, mass of iron(III) oxide = 0.021 moles × 159.69 g/mol = 3.3535 g