<u>Answer:</u> The mass of nonahydrate iron (III) nitrate is 16.2 g
<u>Explanation:</u>
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)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
Molarity of
= 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](https://tex.z-dn.net/?f=0.0200M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DFe%28NO_3%29_3%7D%7B2L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20%7DFe%28NO_3%29_3%3D%280.0200mol%2FL%5Ctimes%202L%29%3D0.04mol)
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](https://tex.z-dn.net/?f=Fe%28NO_3%29_3.9H_2O%5Crightarrow%20Fe%28NO_3%29_3%2B9H_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 =
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}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
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](https://tex.z-dn.net/?f=0.04mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nonahydrate%20iron%20%28III%29%20nitrate%7D%7D%7B404.0g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20nonahydrate%20iron%20%28III%29%20nitrate%7D%3D%280.04mol%5Ctimes%20404.0g%2Fmol%29%3D16.2g)
Hence, the mass of nonahydrate iron (III) nitrate is 16.2 g