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)
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Ответ:
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
Ответ:
the answer is 1.75 i just took the test