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    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.

    Thanks a lot!!!!
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    A sample of solid ethanedioic acid ( \text{H}_2 \text{C}_2 \text{O}_4 \cdot 2\text{H}_2\text{O} ) has been contaminated with potassium ethanedioate ( \text{K}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2 \text{O} ). 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  \text{KMnO}_4 solution, requiring 24.85 cm3. Calculate  x .
     \text{C}_2\text{O}_4 ^{2-} 2\text{CO}_2 +  \text{e}^- .

    You will have to think very carefully about redox reactions and acid-base reactions to answer this question.
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    (Original post by B_9710)
    A sample of solid ethanedioic acid ( \text{H}_2 \text{C}_2 \text{O}_4 \cdot 2\text{H}_2\text{O} ) has been contaminated with potassium ethanedioate ( \text{K}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2 \text{O} ). 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  \text{KMnO}_4 solution, requiring 24.85 cm3. Calculate  x .
     \text{C}_2\text{O}_4 ^{2-} 2\text{CO}_2 +  \text{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?
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    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?
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    (Original post by B_9710)
    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?
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    (Original post by HiThere8964)
    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
     \text{H}_2 \text{C}_2 \text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2 \text{C}_2 \text{O}_4 + 2\text{H}_2 \text{O} .
    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.
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    (Original post by B_9710)
    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
     \text{H}_2 \text{C}_2 \text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2 \text{C}_2 \text{O}_4 + 2\text{H}_2 \text{O} .
    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!!
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    I once punched a mole and I knocked it clean out.

    Probably wasn't all that hard.
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    (Original post by HiThere8964)
    Thanks a lot for the question and the help I will try and rep you again as soon as I can!!
    No problem
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    I still cant get it can you post a worked solution please,
 
 
 
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