Question

A student made solution #3 using the experimental method in this lab, and measured an absorbance of 0.559. The starting reagents are 2.00 x 10-3 M Fe(NO3)3 and 2.00 x 10-3 M KSCN.

The amount of absorption is proportional to the concentration of

FeSCN2+. This relationship – true for many solutions – is called “Beer’s Law”, and has the simple equation:

A = bc

where “A” is the absorption, “b” is 5174.6 for FeSCN2+ and “c” is molarity

Make Five Solutions: Following the mix chart below, add Fe(NOs)s solution, KSCN solution, and perhaps distilled water into ea

Advance Study Assignment A student made lab, and measured an absorbance of 0.559. The starting reagents are 2.00 x 103 M Fe(N

Make Five Solutions: Following the mix chart below, add Fe(NOs)s solution, KSCN solution, and perhaps distilled water into each of the test tubes. The total solution volume in each test tube must be 10.00 mL. Mix with a stirring rod, cleaning between each test tube. Test Tube number 2 3 4 5 Fe(NO,)s sol'n, in mL 5.00 5.00 5.005.00 5.00 KSCN sol'n, in mL 1.00 2.00 3.00 4.00 5.00 Distilled H2O, in mL 4.03.00 2.001.00 equilibrium V, in mL 10.00 10.0010.00 10.00 10.00
Advance Study Assignment A student made lab, and measured an absorbance of 0.559. The starting reagents are 2.00 x 103 M Fe(NO;)s and 2.00 x 103 M KSCN. 1 . What is the equilibrium molarity of FeSCN2+ in solution #3? solution #3 usingthe experimental method inthis 2. How many moles of FeSCN2 are in the solution at equilibrium? equilibrium moles FeSCN2+ 3. What are the initial number of moles of FeSCN2? initial moles FeSCN2+ = 4. What are the initial number of moles of each of the reactants? initial moles Fe3+ initial moles SCN 5. Complete the following I.C.E. table (using moles) for the reaction (Reaction I) moles of... SCN FeSCN2- Fel Initial Change
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Answer #1

Q1:

A = 0.559 = b×c = 5174.6 × c

c = 1.08×10-4 M

[FeSCN2+] = 1.08×10-4M

Q2:

Moles of FeSCN2+ = molarity × volume( in L)

= 1.08×10-4×10.00×10-3 = 1.08×10-6 mol

Q3:

Initial number of moles of FeSCN2+ = 0

Q4:

Initial moles of Fe3+ = 2.00×10-3 × 5.00×10-3 = 1.00×10-5mol

Initial moles of SCN- = 2.00×10-3×3.00×10-3 = 6.00×10-6 mol

Q5:

Fe3+ SCN- FeSCN2+
Initial 1.00×10-5 6.00×10-6 0
Change 1.08×10-6 1.08×10-6 1.08×10-6
Equilibrium 8.92×10-6 4.92×10-6 1.08×10-6
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