Please calculate the 4 concentrations of the FeScN3 in part A2 and fill out the table, show all work. **Part A-2. Calculation for the Concentrations of the Given Solutions**Fill out the table below and determine the concentration of FeSCN²⁺ in each sample. Remember for Part A, the limiting reactant is SCN⁻.---**Table 1: Concentration of Solutions and Absorbance Values**| Beaker | Moles of SCN⁻ | Moles of FeSCN²⁺ | Concentration of FeSCN²⁺ (M) ||--------|--------------|------------------|-----------------------------|| 1 | | | || 2 | | | || 3 | | | || 4 | | | |*Show sample calculation below.*---**Insert your ALL calculations here:** **I. Objective:**To determine the equilibrium constant for the formation of a complex ion, FeSCN²⁺.\[ \text{Fe}^{3+} \, (aq) + \text{SCN}^- \, (aq) \leftrightarrow \text{FeSCN}^{2+} \, (aq) \]\[ K_f = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^-]} \]**II. Observations:****Part A. Constructing a Beer’s Law Plot**Five solutions are prepared according to the table below. We assume in this first part, that the SCN⁻ determines the concentration of FeSCN²⁺ formed (because Fe³⁺ is in excess and SCN⁻ is limiting). So we can calculate the FeSCN²⁺ concentration just by using the moles of SCN⁻ that we started with.**Data Table 1.**| Beaker Number | 0.200 M Fe(NO)₃ (mL) | 0.0020 M KSCN (mL) | H₂O (mL) ||---------------|----------------------|--------------------|----------|| 1 | 5.0 | 5.0 | 40.0 || 2 | 5.0 | 4.0 | 41.0 || 3 | 5.0 | 3.0 | 42.0 || 4 | 5.0 | 2.0 | 43.0 || 5 (Blank) | Data not needed | Data not needed | Data not needed |The samples are run on a visible spectrometer and an **example** of the data is shown below.**Graph Explanation:**The graph titled "Visible Spectra for 4 FeSCN²⁺ solutions" displays the absorbance on the y-axis and the wavelength (nm) on the x-axis. There are four curves, each representing a different FeSCN²⁺ solution concentration. The x-axis ranges from 400 nm to 600 nm, and absorbance is measured from 0 to 1.2. The curves typically show a peak in

In this experiment, the main reaction that is involved can be shown as Fe3++SCN-→FeSCN2+ Now, it has been mentioned that SCN- is the limiting reagent and obviously, Fe3+ will be excess reagent. The limiting reagent in a reaction is the reagent whose amount is limited and the final amount of product will be decided by it. The excess reagent is the reagent that is present in excess and the final amount of product does not depend on it. Hence, the number of moles of FeSCN2+ will depend upon the moles of SCN- not the moles of Fe3+. The source of Fe3+ is Fe(NO3)3 and the source of SCN- is KSCN. According to the main reaction equation, one mole SCN- produces one mole FeSCN2+, we can write moles of SCN-=moles of FeSCN2+ Again, we can get molar concentration by the following equation: molarity =molesvolumeBeaker - 1 Reagents: 5 ml 0.2 M Fe(NO3)3 5 ml 0.002 M KSCN 40 ml H2O Total volume = 5+5+40=50 ml Moles of SCN-= 0.002×51000=10-5 Moles of FeSCN2+= 10-5 Concentration of FeSCN2+ =10-5 moles0.05 litre=2×10-4 M Beaker - 2 Reagents: 5 ml 0.2 M Fe(NO3)3 4 ml 0.002 M KSCN 41 ml H2O Total volume = 5+4+41=50 ml Moles of SCN-= 0.002×41000=8×10-6 Moles of FeSCN2+= 8×10-6 Concentration of FeSCN2+ =8×10-6 moles0.05 litre=1.6×10-4 M Beaker - 3 Reagents: 5 ml 0.2 M Fe(NO3)3 3 ml 0.002 M KSCN 42 ml H2O Total volume = 5+3+42=50 ml Moles of SCN-= 0.002×31000=6×10-6 Moles of FeSCN2+= 6×10-6 Concentration of FeSCN2+ =6×10-6 moles0.05 litre=1.2×10-4 M Beaker - 4 Reagents: 5 ml 0.2 M Fe(NO3)3 2 ml 0.002 M KSCN 43 ml H2O Total volume = 5+2+43=50 ml Moles of SCN-= 0.002×21000=4×10-6 Moles of FeSCN2+= 4×10-6 Concentration of FeSCN2+ =4×10-6 moles0.05 litre=8×10-5 M Hence, finally, the table will be Beaker Moles of SCN- Moles of FeSCN2+ Concentration of FeSCN2+ (M) 1 10-5 10-5 2×10-4 2 8×10-6 8×10-6 1.6×10-4 3 6×10-6 6×10-6 1.2×10-4 4 4×10-6 4×10-6 8×10-5