In the synthesis of Fe3O4, if 50 ml of a solution containing 0.05 M FeCl24H2O an
ID: 1075119 • Letter: I
Question
In the synthesis of Fe3O4, if 50 ml of a solution containing 0.05 M FeCl24H2O and 0.075 M FeCl36H2O is needed, how many grams of FeCl24H2O and how many grams of FeCl36H2O must be mixed in the 50 ml solution? Answer: 0.497 g of FeCl24H2O and 1.013 g of FeCl36H2O.
How many grams of Fe3O4 would be formed if sufficient NH4OH were added? Answer: 0.434 g of Fe3O4 formed.
In the experiment, it was found that the amount of solid formed was 0.495 g, higher than that obtained from (b). There are several possibilities for the larger amount obtained in the experiment. Give a quantitative explanation for the difference by calculating the amount of the additional phase (for example, FeO or Fe(OH)2).
Explanation / Answer
let the volume of FeCl2.4H2O is y mL so the volume of FeCL3.6H2O= (50-y) mL
Molar mass of FeCl2.4H2O=[55.845+(2×35.5)+(4×18)] g = 198.845 g
1000 mL 1 M solution of FeCl2.4H2O contains 198.845 g FeCl2.4H2O
so,50 mL of 0.05 M solution of FeCl2.4H2O contains [198.845×.0.05×50)/1000] g FeCl2.4H2O=0.4971125 g of FeCl2.4H2O
so ,0.4971125 g of FeCl2.4H2O is needed to make a 50 mL solution.
Molar mass of FeCl3.6H2O=[55.845+(3×35.5)+(6×18)] g = 270.345 g
1000 mL 1 M solution of FeCl3.6H2O contains g FeCl3.6H2O
so,50 mL of 0.05 M solution of FeCl3.6H2O contains [270.345×.0.075×50)/1000] g FeCl3.6H2O=1.01379 g of FeCl3.6H2O
so ,1.01379 g of FeCl3.6H2O is needed to make a 50 mL solution.
Fe2+ + 2Fe3+ + OH- = Fe3O4 + 4H2O
molar mass of Fe3O4 = [(3×55.845)+(4×16)] g =231.535 g
so 1 mol of Fe2+ and 2 moles of Fe3+ form 1 mol of Fe3O4
so 1 mole of FeCl2.4H2O=198.845g of FeCl2.4H2O
1mole of FeCl3..6H2O=270.345g of FeCl3.6H2O
to form Fe3O4 the molar ratio of Fe2+:Fe3+ should be maintained to 1:2