0.5468 g of sodium oxalate, Na_2C_2O_4, is titrated with KMnO_4 solution according to the...


0.5468 g of sodium oxalate, {eq}Na_2C_2O_4 {/eq}, is titrated with {eq}KMnO_4 {/eq} solution according to the following net ionic equation:

{eq}5C_2O_4^{2-}(aq) + 2MnO_4^-(aq) + 16H^+(aq) \rightarrow 10CO_2 (g) + 8H_2O (l) + 2Mn^{2+}(aq) {/eq}

It required 35.43 mL of {eq}KMnO_4 {/eq} solution to reach the endpoint.

(a) What compound is represented by the ion {eq}C_2O_4^{2-} {/eq}?

(b) What compound is represented by the ion {eq}MnO_4^- {/eq}?

(c) What is the molar mass of {eq}Na_2C_2O_4 {/eq}?

(d) How many moles of {eq}Na_2C_2O_4 {/eq} are in 0.5468 g?

(e) What is the mole ratio of {eq}KMnO_4 {/eq} to {eq}Na_2C_2O_4 {/eq} in the balanced equation?

(f) How many moles of {eq}KMnO_4 {/eq} are required to react with 0.5468 g of {eq}Na_2C_2O_4 {/eq}?

(g) How many liters is 35.43 mL?

(h) What is the molarity of the {eq}KMnO_4 {/eq} solution?


In a titration, the solution at known concentration usually in the burette is called the titrant, while the solution of analyte being analyzed is called the titrand.

Answer and Explanation: 1

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Part (a) - {eq}C_2O_4^{2-} {/eq}

The ion {eq}C_2O_4^{2-} {/eq} is called the oxalate ion.

Part (b) - {eq}MnO_4^- {/eq}

The ion {eq}MnO_4^- {/eq}...

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Calculating Reaction Yield and Percentage Yield from a Limiting Reactant


Chapter 9 / Lesson 6

How to calculate the theoretical yield? Learn the definition and formula of percent yield. Use the theoretical yield equation to calculate theoretical yield.

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