Please could someone link me some hard moles equations suitable for an AS Chemistry student. I would appreciate hard moles questions that really make me apply my knowledge.
A sample of solid ethanedioic acid (H2C2O4⋅2H2O) has been contaminated with potassium ethanedioate (K2C2O4⋅xH2O). A 1.780 g sample of this mixture was made up to a 250 cm3 solution with distilled water. A 25 cm3 sample was titrated against 0.100 mol dm-3 sodium hydroxide, requiring 17.35 cm3. Another 25 cm3 sample was acidified with sulphuric acid and titrated against 0.0200 mol dm-3 KMnO4 solution, requiring 24.85 cm3. Calculate x. C2O42− → 2CO2 + e−.
You will have to think very carefully about redox reactions and acid-base reactions to answer this question.
A sample of solid ethanedioic acid (H2C2O4⋅2H2O) has been contaminated with potassium ethanedioate (K2C2O4⋅xH2O). A 1.780 g sample of this mixture was made up to a 250 cm3 solution with distilled water. A 25 cm3 sample was titrated against 0.100 mol dm-3 sodium hydroxide, requiring 17.35 cm3. Another 25 cm3 sample was acidified with sulphuric acid and titrated against 0.0200 mol dm-3 KMnO4 solution, requiring 24.85 cm3. Calculate x. C2O42− → 2CO2 + e−.
You will have to think very carefully about redox reactions and acid-base reactions to answer this question.
Thanks for the question, this is really tough too I have calculated the moles of the titrations and I don't know what I should do next?
Think about the reactions and what it means. The base (NaOH) added will only react with the acid (acid-base) reaction (advice is to write the equation for this) but the KMnO4 will react with the C2O4- ions in both the acid and the impurity (the K2C2O4.xH2O). Perhaps you might want to post your workings?
Think about the reactions and what it means. The base (NaOH) added will only react with the acid (acid-base) reaction (advice is to write the equation for this) but the KMnO4 will react with the C2O4- ions in both the acid and the impurity (the K2C2O4.xH2O). Perhaps you might want to post your workings?
I don't understand what the equation should be. do you mean K2C2O4.XH2O + KMnO4 and H2C2O4.2H2O + NaOH if so what would the products be?
I don't understand what the equation should be. do you mean K2C2O4.XH2O + KMnO4 and H2C2O4.2H2O + NaOH if so what would the products be?
For the equations forget about the .H2O (called water of crystallisation I believe) as its not important for the reactions. The acid-base reaction between NaOH and the acid H2C2O4 (forgetting the .H2O now) is H2C2O4+2NaOH→Na2C2O4+2H2O. Use this to work out the moles of H2C2O4.2H2O in the 25cm3 sample. Use the half equation given to write a redox equation between MnO4- and C2O4 2-. Then see where you can take it from there.
For the equations forget about the .H2O (called water of crystallisation I believe) as its not important for the reactions. The acid-base reaction between NaOH and the acid H2C2O4 (forgetting the .H2O now) is H2C2O4+2NaOH→Na2C2O4+2H2O. Use this to work out the moles of H2C2O4.2H2O in the 25cm3 sample. Use the half equation given to write a redox equation between MnO4- and C2O4 2-. Then see where you can take it from there.
Thanks a lot for the question and the help I will try and rep you again as soon as I can!!