Asked by Kevin
If 7.83 g FeCl3(s) is dissolved in enough water to make 2.50 L of solution, what is the concentration of Fe3+ in the solution?
Answers
Answered by
John
First you have to get from grams of FeCl3 to moles by dividing the grams by the molar mass of FeCl3, 162.21.
7.83 / 162.21 = .0483 mols
Second, you have to convert from moles of FeCl3 to moles of Fe. With this compound, for every mole of FeCl3 there is one mole of Fe.
.0483 mols FeCl3 = .0483 mols Fe
Lastly, find molarity by dividing moles by liters.
.0483 mols / 2.50 L = .0193 M
7.83 / 162.21 = .0483 mols
Second, you have to convert from moles of FeCl3 to moles of Fe. With this compound, for every mole of FeCl3 there is one mole of Fe.
.0483 mols FeCl3 = .0483 mols Fe
Lastly, find molarity by dividing moles by liters.
.0483 mols / 2.50 L = .0193 M
Answered by
DrBob222
In what units?
7.83 g FeCl3/2.50 L = 3.13 g/L or
3.13 x (55.85/162.2) = 1.08 g Fe^+3/L.
7.83 g FeCl3/2.50 L = 3.13 g/L or
3.13 x (55.85/162.2) = 1.08 g Fe^+3/L.
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