Question

(b) If a 1.50−g sample of plant matter requires 39.0 mL of 0.0500 M KMnO₄ solution to reach the equivalence point, what is the percent by mass of H₂C₂O₄ in the sample

Answer 1

H2C2O4 (aq) + MnO4 (aq) -----> CO2 + Mn+2 (aq) The half reactions are Oxidation H2C204 ----->CO2 Reduction MnO4-------> Mn+2 H2C204 -----> 2 CO2 Balance atoms other than H and o MnO4-------> Mn+2 Mn0,-------> Mn+2+ 4 H,O Balance Oxygen: H2C204 -----> 2 CO2 (Add H2O on less side) Balance Hydrogen: H2C204 -----> 2 CO2 + 2 H+ (Add Ht on less side) Mn0, +8 H+ -----> Mn+2+ 4 H2O Balance charge: H2C204 -----> 2 CO2 + 2H+ + 2 e 5H2C204 ------> 10 CO2 + 10 H+ + 10e MnO4 + 8 H+ + 5 e ------> Mn+2 + 4H2O 2 MnO4 +16 H+ + 10 e ------> 2 Mn+2+ 8 H2O Combine: 5 H,C204 + 2 MnO4 + 16 H+ + 10 e ---------> 10 CO2 + 10 H+ + 10e + 2 Mn+2 + 8 H2O 5 H2C2O4 + 2 MnO4 +6 H+ ---------> 10 CO2 + 2 Mn+2+ 8 H20 The balanced equation is 5 H2C2O4 (aq) + 2 MnO,- (aq) + 6 H+ (aq) ---------> 10 CO2 (g) + 2 Mn+2 (aq) +8 H20 (1)

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The balanced equation is 5 H20204 (aq) + 2 MnO,- (aq) + 6 H+ (aq) ---------> 10 CO2 (g) + 2 Mn+2 (aq) + 8 H20 (1) Number of moles of MnO4 = M*V = 0.0500 M * 39.0 mL = 1.95 mmol = 0.00195 mol According to balanced equation, 2 moles of MnO4 reacts with 5 moles of H2C204 So, 0.00195 mol of Mno, would react with 0.00195 mol * 5/2 = 0.004875 mol of H,C,04 Number of moles of H2C204 = 0.004875 mol Mass of H20204 = moles * mol.wt. = 0.004875 mol * 90.03 g/mol = 0.438896 g Percent by mass = mass of H2C204/mass of sample * 100 Percent by mass = 0.438896 /1.50 * 100 = 29.26 % percent by mass of H2C2O4 in the sample = 29.26 %