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?
You add the zeros.
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 ..??.......??.......??...........??
Your tasks.
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.
To determine the concentration of all ions present after the reaction is complete, we first need to write out the balanced equation for the reaction between sodium carbonate and iron (III) nitrate:
Na2CO3 + 2Fe(NO3)3 → 2NaNO3 + Fe2(CO3)3
From the balanced equation, we can see that sodium carbonate reacts with iron (III) nitrate to form sodium nitrate and iron (III) carbonate as an insoluble precipitate.
Now, let's determine the moles of each reactant by using the given volumes and concentrations.
For the sodium carbonate solution:
Volume = 1.000 L
Concentration = 0.750 M
Moles of sodium carbonate = Volume x Concentration
= 1.000 L x 0.750 mol/L
= 0.750 mol
For the iron (III) nitrate solution:
Volume = 0.5000 L
Concentration = 1.50 M
Moles of iron (III) nitrate = Volume x Concentration
= 0.5000 L x 1.50 mol/L
= 0.750 mol
According to the balanced equation, the stoichiometry is 1:2 between sodium carbonate and iron (III) nitrate. This means that 1 mole of sodium carbonate reacts with 2 moles of iron (III) nitrate.
Since we have equal moles of both reactants, the limiting reagent in this case is sodium carbonate, as it will be completely consumed in the reaction, while there will be excess iron (III) nitrate.
Therefore, after the reaction is complete, all 0.750 moles of sodium carbonate will have reacted, producing 0.750 moles of sodium nitrate and 0.375 moles of iron (III) carbonate.
Now, let's calculate the concentration of each ion in the final solution:
Since 1 mole of sodium carbonate produces 2 moles of sodium ions, the concentration of sodium ions is:
Concentration of sodium ions = (moles of sodium nitrate + moles of sodium carbonate) / total volume
Concentration of sodium ions = (0.750 mol + 0.750 mol) / (1.000 L + 0.5000 L)
= 1.500 mol / 1.500 L
= 1.000 M
Since 1 mole of iron (III) nitrate produces 0 moles of iron (III) ions, and 1 mole of iron (III) carbonate produces 2 moles of iron (III) ions, the concentration of iron (III) ions is:
Concentration of iron (III) ions = (2 × moles of iron (III) carbonate) / total volume
Concentration of iron (III) ions = (2 × 0.375 mol) / (1.000 L + 0.5000 L)
= 0.750 mol / 1.500 L
= 0.500 M
Therefore, the concentration of sodium ions in the final solution is 1.000 M, and the concentration of iron (III) ions is 0.500 M.