1.) Given the following balanced chemical equation for the titration of oxalate

Question

1.) Given the following balanced chemical equation for the titration of oxalate ion with permanganate ion:                 

6H+ + 2MnO4- + 5C2O42- --> 2Mn+  + 10CO2 + 8H2O

                    A student titrated a 0.164 g sample of crystals with 20.50 mL 0.0300 M KMnO4. Calculate the mass of oxalate ion in the sample?                 

                    Show Your Work!!!
                

Explanation / Answer

16 H+ + 2 MnO4- + 5 C2O42- => 2 Mn2+ + 10 CO2 + 8 H2O

Moles of MnO4- = volume x concentration of KMnO4

= 20.50/1000 x 0.0300 = 0.000615 mol


Moles of C2O42- = 5/2 x moles of MnO4-

= 5/2 x 0.000615 = 0.0015375 mol


Mass of C2O42- = moles x molar mass of C2O42-

= 0.0015375 x 88.02

= 0.1353 g = 0.135 g



(Note: Mass percent of C2O42- = 0.1353/0.164 x 100% = 82.5%)

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