Question

In a volumetric analysis experiment, a solution of sodium oxalate (Na2C2O4) in acidic solution is titrated with a solution of potassium permanganate (KMnO4) according to the following balanced chemical equation: 2KMnO4(aq) + 8H2SO4(aq) + 5Na2C2O4(aq) → 2MnSO4(aq) + 8H2O(l) + 10CO2(g) + 5Na2SO4(aq) + K2SO4(aq). It required 40.0 mL of 0.0335 M KMnO4 to reach the endpoint. What mass of Na2C2O4 was present initially? a) 11.2 g b) 0.0718 g c) 0.448 g d) 1.79 g

Name: in a volumetric analysis experiment a solution of sodium oxalate na2c2o4 in acidic solution is titrated with a solution of potassium permanganate kmno4 according to the following balanced ch 18939
Uploaded: 2022-08-10T09:49:59-08:00
Duration: 1 min 49 s
Channel: Ma Ednelyn Lim
Description: in a volumetric analysis experiment a solution of sodium oxalate na2c2o4 in acidic solution is titrated with a solution of potassium permanganate kmno4 according to the following balanced ch 18939

          In a volumetric analysis experiment, a solution of sodium oxalate (Na2C2O4) in acidic solution is titrated with a solution of potassium permanganate (KMnO4) according to the following balanced chemical equation: 2KMnO4(aq) + 8H2SO4(aq) + 5Na2C2O4(aq) → 2MnSO4(aq) + 8H2O(l) + 10CO2(g) + 5Na2SO4(aq) + K2SO4(aq). It required 40.0 mL of 0.0335 M KMnO4 to reach the endpoint. What mass of Na2C2O4 was present initially?
a) 11.2 g
b) 0.0718 g
c) 0.448 g
d) 1.79 g

Added by Brian A.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Ma Ednelyn Lim

08/10/2022

Step 1

The KMnO4 solution has a concentration of 0.0335 M and a volume of 40.0 mL, which is 0.0400 L. Calculation: moles of KMnO4 = 0.0335 M × 0.0400 L = 0.00134 moles. Show more…

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In a volumetric analysis experiment, a solution of sodium oxalate (Na2C2O4) in acidic solution is titrated with a solution of potassium permanganate (KMnO4) according to the following balanced chemical equation: 2KMnO4(aq) + 8H2SO4(aq) + 5Na2C2O4(aq) → 2MnSO4(aq) + 8H2O(l) + 10CO2(g) + 5Na2SO4(aq) + K2SO4(aq). It required 40.0 mL of 0.0335 M KMnO4 to reach the endpoint. What mass of Na2C2O4 was present initially? a) 11.2 g b) 0.0718 g c) 0.448 g d) 1.79 g