Answer :
Answer: The minimum mass of iron (II) nitrate that must be added is 0.188 grams
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of phosphate solution = 0.0699 M
Volume of solution = 10 mL = 0.010 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
[tex]0.0699M=\frac{\text{Moles of phosphate solution}}{0.010L}\\\\\text{Moles of phosphate solution}=(0.0699mol/L\times 0.010L)=6.99\times 10^{-4}mol[/tex]
The given chemical equation follows:
[tex]2PO_4^{3-}(aq.)+3Fe(NO_3)_2(aq.)\rightarrow Fe_3(PO_4)_2(s)+6NO_3^-(aq.)[/tex]
By Stoichiometry of the reaction:
2 moles of phosphate solution reacts with 3 moles of iron (II) nitrate
So, [tex]6.99\times 10^{-4}mol[/tex] of phosphate solution will react with = [tex]\frac{3}{2}\times 6.99\times 10^{-4}=1.049\times 10^{-3}mol[/tex] of iron (II) nitrate
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of iron (II) nitrate = 180 g/mol
Moles of iron (II) nitrate = [tex]1.049\times 10^{-3}[/tex] moles
Putting values in above equation, we get:
[tex]1.049\times 10^{-3}mol=\frac{\text{Mass of iron (II) nitrate}}{180g/mol}\\\\\text{Mass of iron (II) nitrate}=(1.049\times 10^{-3}mol\times 180g/mol)=0.188g[/tex]
Hence, the minimum mass of iron (II) nitrate that must be added is 0.188 grams