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

Answer:

3.3535 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

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 :

$Moles=0.404 \times {52.0\times 10^{-3}}\ moles$

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