Question

In reality, a hydrate of iron(III) nitrate had to be used, not the anhydrous salt. As you may guess, some of the hydrate’s mass is water, and some is iron(III) nitrate. How many grams of Fe(NO3)3•9H2O needed to be dissolved in water to make 2 L of 0.0020 M Fe(NO3)3? Molecular weight of the nonahydrate is 404.0 g/mol. Hint: try to set up an equation using x, and solving it. Assume that the density of your solution is 1.000 g/mL.

Answer 1

Answer: The mass of nonahydrate iron (III) nitrate is 16.2 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of $Fe(NO_3)_3$ = 0.0020 M

Volume of solution = 2 L

Putting values in above equation, we get:

$0.0200M=\frac{\text{Moles of }Fe(NO_3)_3}{2L}\\\\\text{Moles of }Fe(NO_3)_3=(0.0200mol/L\times 2L)=0.04mol$

The chemical equation for the decomposition of hydrated iron (III) nitrate follows:

$Fe(NO_3)_3.9H_2O\rightarrow Fe(NO_3)_3+9H_2O$

By Stoichiometry of the reaction:

1 mole of iron (III) nitrate is produced from 1 mole of hydrated iron (III) nitrate

So, 0.04 moles of iron (III) nitrate will be produced from = $\frac{1}{1}\times 0.04=0.04mol$ of hydrated iron (III) nitrate

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of nonahydrate iron (III) nitrate = 404.0 g/mol

Moles of nonahydrate iron (III) nitrate = 0.04 moles

Putting values in above equation, we get:

$0.04mol=\frac{\text{Mass of nonahydrate iron (III) nitrate}}{404.0g/mol}\\\\\text{Mass of nonahydrate iron (III) nitrate}=(0.04mol\times 404.0g/mol)=16.2g$

Hence, the mass of nonahydrate iron (III) nitrate is 16.2 g