the following equation describes the oxidation of an acidifiedsolution of sodium

Question

the following equation describes the oxidation of an acidifiedsolution of sodium oxalate (Na2C2O4) by potassium permanganate(kMnO4); 5C2O4-2+ 2MnO4-+16H2SO4-->10CO2+2Mn2+ +16HSO4-+8H2O what volume of 0.127M sodium oxalate in sulphuric acid willreact with 3.57g of potassium permanganate? the following equation describes the oxidation of an acidifiedsolution of sodium oxalate (Na2C2O4) by potassium permanganate(kMnO4); 5C2O4-2+ 2MnO4-+16H2SO4-->10CO2+2Mn2+ +16HSO4-+8H2O what volume of 0.127M sodium oxalate in sulphuric acid willreact with 3.57g of potassium permanganate?

Explanation / Answer

Moles of potassium permangnate = Mass / molar mass
                                                   =3.57 g / 158.03 g/mol                                                    =0.0225 moles 5C2O4-2+ 2MnO4-+16H2SO4-->10CO2+2Mn2+ +16HSO4-+8H2O In the reaction, we can see 5 moles of oxalate reacting with 2moles of permangnate. So moles of oxalate required = 0.0225 * 5 / 2 = 0.05625mol Volume of sodium oxalate = moles / molarity                                         =0.05625 mol / 0.127 M                                         =0.44291 L or 442.91 mL

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.