Determine the amount concentration of the iron ions and sulfate ions in a 0.620 mol/L solution of iron(III) sulfate.
Including the balanced chemical equation.
There is no equation to balance.
Fe2(SO4)3 ==> 2Fe^+3 + 3SO4^-2
So I look at 0.620 M Fe2(SO4)3.
So Fe^+3 must be twice that
and SO4^-2 must be three times that. right?
Don't make problems any harder than they need to be.;-)
thank you
To determine the amount concentration of iron ions (Fe3+) and sulfate ions (SO42-) in a 0.620 mol/L solution of iron(III) sulfate (Fe2(SO4)3), we need to consider the balanced chemical equation for the dissociation of iron(III) sulfate in water.
The balanced chemical equation for the dissociation of iron(III) sulfate is:
Fe2(SO4)3 (s) -> 2 Fe3+ (aq) + 3 SO42- (aq)
From the equation, we can see that one mole of iron(III) sulfate dissociates to produce 2 moles of Fe3+ ions and 3 moles of SO42- ions.
Given that the concentration of the iron(III) sulfate solution is 0.620 mol/L, we can use this information to determine the amount concentration of the iron and sulfate ions.
For iron ions (Fe3+):
The concentration of iron(III) sulfate is 0.620 mol/L, and each mole of iron(III) sulfate yields 2 moles of Fe3+ ions.
Therefore, the amount concentration of Fe3+ ions is 2 * 0.620 = 1.240 mol/L.
For sulfate ions (SO42-):
The concentration of iron(III) sulfate is 0.620 mol/L, and each mole of iron(III) sulfate yields 3 moles of SO42- ions.
Therefore, the amount concentration of SO42- ions is 3 * 0.620 = 1.860 mol/L.
To summarize:
- The amount concentration of iron ions (Fe3+) is 1.240 mol/L.
- The amount concentration of sulfate ions (SO42-) is 1.860 mol/L.
To determine the amount concentration of iron ions (Fe3+) and sulfate ions (SO4 2-) in a solution of iron(III) sulfate (Fe2(SO4)3), we need to start by finding the number of moles of each ion present in the solution.
The balanced chemical equation for the dissolution of iron(III) sulfate is:
Fe2(SO4)3(s) → 2 Fe3+(aq) + 3 SO4 2-(aq)
From the equation, we can see that for every mole of Fe2(SO4)3 dissolved, we get 2 moles of Fe3+ ions and 3 moles of SO4 2- ions.
Given that the solution has a concentration of 0.620 mol/L, it means that for every liter of the solution, we have 0.620 moles of Fe2(SO4)3.
To determine the amount concentration of the iron ions (Fe3+), we multiply the concentration of Fe2(SO4)3 by the mole ratio between Fe3+ and Fe2(SO4)3:
Amount concentration of Fe3+ = 0.620 mol/L × 2 = 1.24 mol/L
Similarly, to determine the amount concentration of sulfate ions (SO4 2-), we multiply the concentration of Fe2(SO4)3 by the mole ratio between SO4 2- and Fe2(SO4)3:
Amount concentration of SO4 2- = 0.620 mol/L × 3 = 1.86 mol/L
Therefore, in a 0.620 mol/L solution of iron(III) sulfate, the amount concentration of iron ions (Fe3+) is 1.24 mol/L, and the amount concentration of sulfate ions (SO4 2-) is 1.86 mol/L.