Table 1: Standardization of KMnO 4 Trial 1 Trial 2 Vol Na 2 C 2 O 4 (mL) 10 mL 1
ID: 503971 • Letter: T
Question
Table 1: Standardization of KMnO4
Trial 1
Trial 2
Vol Na2C2O4 (mL)
10 mL
10 mL
Initial Buret (mL)
0 mL
13.8 mL
Final Buret (mL)
13.8 mL
25 mL
Vol KMnO4 (mL)
13.8 mL
11.2 mL
Moles C2O42-(mol)
0.0005 mol
0.0005 mol
Moles KMnO4(mol)
0.0002 mol
0.0002 mol
Molarity KMnO4(M)
0.01449 M
0.01785 M
Average Molarity (M)
0.01617 M
Table 2: Determination of Iron (II) Percentage
Unknown:ALASKA
Trial 1
Trial 2
Mass of sample (g)
0.3686 g
0.3045 g
Initial Buret (mL)
25 mL
0 mL
Final Buret (mL)
50 mL
25 mL
Vol KMnO4 (mL)
25mL
25 mL
Moles KMnO4
Moles Fe(II)
Mass of Fe in your sample
% Fe in Sample
Average
Theoretical
Error
PLEASE PLEASE PLEASE help me find the values that are missing! im very confused and need to write a lab report on this by tonight!
Trial 1
Trial 2
Vol Na2C2O4 (mL)
10 mL
10 mL
Initial Buret (mL)
0 mL
13.8 mL
Final Buret (mL)
13.8 mL
25 mL
Vol KMnO4 (mL)
13.8 mL
11.2 mL
Moles C2O42-(mol)
0.0005 mol
0.0005 mol
Moles KMnO4(mol)
0.0002 mol
0.0002 mol
Molarity KMnO4(M)
0.01449 M
0.01785 M
Average Molarity (M)
0.01617 M
Explanation / Answer
Trial (1):
moles of KMnO4 = MV = 0.01449 *25*10^-3 = 0.00036225
moles of Fe(II)
relate via
Fe2+(aq) + MnO4–(aq) --> Fe3+(aq) + Mn2+(aq)
so
1:1 ratio, then moles of Fe(II)= 0.00036225 mol
mass of Fe = mol*MW = (0.00036225)(55.85) = 0.020231 g of Fe in sample
%Fe sample = mass of Fe/ Total mass *!00% = 0.020231 / 0.3686 *100 = 5.488%
Theoretical will require actual values, which are not included, so error can't be calculated
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