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)

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:

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:

The given chemical equation follows:

By Stoichiometry of the reaction:

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

So, of phosphate solution will react with = of iron (II) nitrate

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

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

Moles of iron (II) nitrate = moles

Putting values in above equation, we get:

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

idk i’m struggling too