In Part A of this experiment, you prepare 5 solutions of FeSCN2* [one of them is just a blank solution] according to the reaction below. Fe3+ (aq) + SCN- (aq) → FESCN²+ (aq) We assume that the starting SCN determines the concentration of formed (because Fe3+ is in excess and SCN- is limiting). Calculate the concentration of FeSCN2+ that forms for each of the solution (Beakers 1- 4) and fill out the table below. Show your calculations beneath the table. Beaker 0.200 M 0.0020 M Concentration of Number Fe(NO3)3 (mL] KSCN (mL] H20 [mL] FESCN2+ (M) 1 5.0 5.0 40.0 0.0002 2 5.0 4.0 41.0 0.00016 5.0 3.0 42.0 0.00012 4 5.0 2.0 43.0 0.00008 5 (blank] 5.0 0.0 45.0 181 Show your work for question 1 here: Total volume of all beakers = 50 mL MSCN x V SCN [FESCN2*] = [SCN') = values ok but, you should NOT Use M.VI=M2V2 Total Volume Beaker 1: 0.002 x 5 (FESCN2*] = 50 = 0.0002 M Beaker 2: 0.002 x 4 (FESCN2*] = = 0.00016 M 50 Beaker 3: Show mol ratio 0.002 x 3 (FESCN2"] = = 0.00012M 50 Beaker 4: 0.002 x 4 [FESCN2*] = = 0.00008M 50

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In Part A of this experiment, you prepare 5 solutions of FeSCN2* [one of them is just a blank
solution] according to the reaction below.
Fe3+ (aq) + SCN¯(aq) –
FESCN²+ (aq)
We assume that the starting SCN determines the concentration of formed (because Fe3+ is in
excess and SCN¯ is limiting). Calculate the concentration of FeSCN2* that forms for each of the
solution (Beakers 1- 4) and fill out the table below. Show your calculations beneath the table.
Beaker
0.200 M
0.0020 M
Concentration of
Number
Fe(NO3)3 [mL]
KSCN [mL]
H20 [mL]
FESCN2* (M)
5.0
5.0
40.0
0.0002
2
5.0
4.0
41.0
0.00016
5.0
3.0
42.0
0.00012
4
5.0
2.0
43.0
0.00008
5 (blank]
5.0
0.0
45.0
181
Show your work for question 1 here:
Total volume of all beakers = 50 mL
MSCN x VSCN
[FESCN2*] = [SCN'] =;
values ok but,
you Should
NOT
%3D
Total Volume
Beaker 1:
0.002 x 5
(FESCN2*) =
50
= 0.0002 M
Beaker 2:
use
0.002 x 4
(FESCN2*] =
= 0.00016 M
MIVI=M2V2
Show mol ratio
50
Beaker 3:
0.002 x 3
(FESCN2*] =
50
= 0.00012 M
Beaker 4:
0.002 x 4
(FESCN2*] =
= 0.00008 M
50
Transcribed Image Text:In Part A of this experiment, you prepare 5 solutions of FeSCN2* [one of them is just a blank solution] according to the reaction below. Fe3+ (aq) + SCN¯(aq) – FESCN²+ (aq) We assume that the starting SCN determines the concentration of formed (because Fe3+ is in excess and SCN¯ is limiting). Calculate the concentration of FeSCN2* that forms for each of the solution (Beakers 1- 4) and fill out the table below. Show your calculations beneath the table. Beaker 0.200 M 0.0020 M Concentration of Number Fe(NO3)3 [mL] KSCN [mL] H20 [mL] FESCN2* (M) 5.0 5.0 40.0 0.0002 2 5.0 4.0 41.0 0.00016 5.0 3.0 42.0 0.00012 4 5.0 2.0 43.0 0.00008 5 (blank] 5.0 0.0 45.0 181 Show your work for question 1 here: Total volume of all beakers = 50 mL MSCN x VSCN [FESCN2*] = [SCN'] =; values ok but, you Should NOT %3D Total Volume Beaker 1: 0.002 x 5 (FESCN2*) = 50 = 0.0002 M Beaker 2: use 0.002 x 4 (FESCN2*] = = 0.00016 M MIVI=M2V2 Show mol ratio 50 Beaker 3: 0.002 x 3 (FESCN2*] = 50 = 0.00012 M Beaker 4: 0.002 x 4 (FESCN2*] = = 0.00008 M 50
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