Oxalic acid (H2C2O4) reacts with the oxidizing agent KMnO4 according to the following reaction:?
Oxalic acid (H2C2O4) reacts with the oxidizing agent KMnO4 according to the following reaction:
6H+ + 5H2C2O4 + 2MnO4- --> 2Mn2+ + 10CO2 + 8H2O
Suppose a 8.42 g sample of green leaves that contain oxalic acid requires 33.45 mL of 0.025 M KMnO4 for titration to the equivalence point. What is the mass percent of oxalic acid in the leaves?
Answer:
Moles KMnO4 = 33.45 x 0.025 /1000 = 0.000836
The ratio between MnO4- and H2C2O4 is 2 : 5
2 : 5 = 0.000836 : x
x = 0.00209 ( moles H2C2O4 )
Molecular weight H2C2O4 = 90 g/mol
0.00209 mol x 90 g/mol = 0.188 g
0.188 : 8.42 = x : 100
x = 2.23 %
First determine the number of moles of KMnO4 you have
0.025molesKMnO4 x 0.03345L=
..1Liter.KMnO4
=8.6x10-5moles KMnO4
Considering the mole to mole ratio is 5 moles Oxalic acid to 2 moles of the permanganate ion then you have:
8.6X10-5*(5/2)= 2.15x10-4 moles Oxalic acid
Now convert to grams using the molar mass of oxalic acid
2.15x10-4moles OA x 90.0354g OA=
......... mole OA
(1.9x10-2g Oxalic acid/8.42)*100=0.23%
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6H+ + 5H2C2O4 + 2MnO4- --> 2Mn2+ + 10CO2 + 8H2O
Suppose a 8.42 g sample of green leaves that contain oxalic acid requires 33.45 mL of 0.025 M KMnO4 for titration to the equivalence point. What is the mass percent of oxalic acid in the leaves?
Answer:
Moles KMnO4 = 33.45 x 0.025 /1000 = 0.000836
The ratio between MnO4- and H2C2O4 is 2 : 5
2 : 5 = 0.000836 : x
x = 0.00209 ( moles H2C2O4 )
Molecular weight H2C2O4 = 90 g/mol
0.00209 mol x 90 g/mol = 0.188 g
0.188 : 8.42 = x : 100
x = 2.23 %
First determine the number of moles of KMnO4 you have
0.025molesKMnO4 x 0.03345L=
..1Liter.KMnO4
=8.6x10-5moles KMnO4
Considering the mole to mole ratio is 5 moles Oxalic acid to 2 moles of the permanganate ion then you have:
8.6X10-5*(5/2)= 2.15x10-4 moles Oxalic acid
Now convert to grams using the molar mass of oxalic acid
2.15x10-4moles OA x 90.0354g OA=
......... mole OA
(1.9x10-2g Oxalic acid/8.42)*100=0.23%
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