# chemistry

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an insoluble precipitate was formed when a solution of 1.000 L of 0.750 M sodium carbonate (Na2CO3) was mixed with .5000 L of 1.50 M solution of iron (III) nitrate (Fe(NO3)3). what is the concentration of all ions present after the reaction is complete?

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moles Na2CO3 = M x L = 1 L x 0.750 = 0.75
moles Fe(NO3)3 = 0.5X1.5 = 0.75
I = initial
C = change
E = equilibrium
3Na2CO3 + 2Fe(NO3)3 > 6NaNO3 + Fe2(CO3)3
I 0.75.....0.75............0........0

Determine limiting reagent. I do it by working two stoichiometry problems, one for the Na2CO3 to determine how much Fe2CO3 is formed and the other for Fe(NO3)3 to determine Fe2(CO3)3 formed, then the smaller of the two is the limiting reagent. Use the coefficients in the balanced equation to do that.
0.75moles Na2CO3 x (1 mol Fe2(NO3)3/3 moles Na2CO3 = 0.75*(1/3) = 0.25 moles Fe2(CO3)3 possibly formed.
0.75 moles Fe(NO3)3 x (1 mole Fe2(CO3)3/2 moles Fe(NO3)3) = 0.75*(1/2) = 0.375 moles Fe2(CO3)3 possibly formed.
Therefore, Na2CO3 is the limiting reagent. You can now finish the ICE chart started above.
3Na2CO3 + 2Fe(NO3)3 > 6NaNO3 + Fe2(CO3)3
I 0.75.....0.75............0........0
C -0.75.....??.........??........0.250
E ..??.......??.......??...........??
1. You know ALL of the Na2CO3 is used.
2. Convert 0.75 moles Na2CO3 to moles Fe2(CO3)3 and write under. We did that above so I've written 0.250 there.
3. In a similar manner, determine moles NaNO3 of the product and moles Fe(NO3)3 used.
4. Now add row 1 to row 2 to arrive at E, the row 3.

Almost finally, M = moles/L for each.

Finally, since Fe2(CO3)3 is not soluble, you must apply Ksp to that to determine the concn of Fe(III) and CO3^-2. Don't forget that you have a common ion from the Fe(NO3)3 and that will reduce the solubility of Fe2(CO3)3.