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Hard Moles Questions

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!!!!

Reply 1

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.

(edited 7 years ago)

Reply 2

HiThere8964

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}

\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 :biggrin:

I have calculated the moles of the titrations and I don't know what I should do next?

Reply 3

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?

Reply 4

HiThere8964

Original post by B_9710

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?

Reply 5

B_9710

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.

Reply 6

HiThere8964

Original post by B_9710

\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.