Question

What minimum mass of iron (II) nitrate must be added to 10.0 of a 0.0699 M phosphate solution in order to completely precipitate all of the phosphate as solid iron (II) phosphate? 2PO43–(aq) + 3Fe(NO3)2(aq) → Fe3(PO4)2(s) + 6NO3–(aq)

Answer 1

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:

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

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:

$0.0699M=\frac{\text{Moles of phosphate solution}}{0.010L}\\\\\text{Moles of phosphate solution}=(0.0699mol/L\times 0.010L)=6.99\times 10^{-4}mol$

The given chemical equation follows:

$2PO_4^{3-}(aq.)+3Fe(NO_3)_2(aq.)\rightarrow Fe_3(PO_4)_2(s)+6NO_3^-(aq.)$

By Stoichiometry of the reaction:

2 moles of phosphate solution reacts with 3 moles of iron (II) nitrate

So, $6.99\times 10^{-4}mol$ of phosphate solution will react with = $\frac{3}{2}\times 6.99\times 10^{-4}=1.049\times 10^{-3}mol$ of iron (II) nitrate

To calculate the number of moles, we use the equation:

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

Molar mass of iron (II) nitrate = 180 g/mol

Moles of iron (II) nitrate = $1.049\times 10^{-3}$ moles

Putting values in above equation, we get:

$1.049\times 10^{-3}mol=\frac{\text{Mass of iron (II) nitrate}}{180g/mol}\\\\\text{Mass of iron (II) nitrate}=(1.049\times 10^{-3}mol\times 180g/mol)=0.188g$

Hence, the minimum mass of iron (II) nitrate that must be added is 0.188 grams