You are Here: Home

# AS Help - Couple Questions can't figure out even with the mark scheme

Announcements Posted on
TSR's new app is coming! Sign up here to try it first >> 17-10-2016
1. Could anyone help me with 2 questions on past papers I just can't figure out. They are both from AQA CHEM3T/P10 from 2010.

The first is on atom economy/yield. The question states: "A second method by which ethanoic acid is synthesised involves the oxidative fermentation of ethanol in the presence of bacteria. The equation representing this reaction is given below.
C2H5OH + O2 -> CH3COOH + H2O
In a small scale experiment using this second method it was found that 23g of ethanol produced only 4.54g of ethanoic acid. Calculate the percentage yield for this experiment."

The mark scheme then says: "1st mark - 23.0g of ethanol produces 30.0g ethanoic acid. 2nd mark - 15.1% (4.54 x 100 / 30)."
I'm fine with the second mark calculating yield, but how do you work out the theoretical production as 30g?? I don't get what I'm missing...

Solved: I was calculating the Mr for Ethanol wrong and it was screwing me up, but now I realise that you use the mass of Ethanol divided by the Mr to work out the moles, then times the moles by the Mr for ethanoic acid and it gives you the theoretical yield of 30g. Doy!

The second question is a simple ideal gas equation calculation. It states: "An excess pf sodium hydrogencarbonate was added to a 10.0cm3 sample of a 15.0 moldm-3 solution of ethanoic acid. Use the ideal gas equation and your answer from question 11 (a) to calculate the volume of carbon dioxide gas that would be formed at 20 degrees C and 1.00 x 10^5 Pa."
The question in 11(a) was to use the equation:
2CH3COOH + Na2CO3 -> 2CH3COONa + H2O +CO2
to calclulate moles of ethanoic acid in 10.0cm3 of 15.0 moldm-3 solution (which comes to 0.150mol).
Using the equation, I halved this amount to 0.075, as there are half the moles of Carbon Dioxide produced. However, in the mark scheme, it uses the value 0.150mol in the ideal gas equation, the same as that for ethanoic acid. What am I missing here?

Thanks for any help and I massively appreciate anything.
2. (Original post by Bacardigan)
Could anyone help me with 2 questions on past papers I just can't figure out. They are both from AQA CHEM3T/P10 from 2010.

The first is on atom economy/yield. The question states: "A second method by which ethanoic acid is synthesised involves the oxidative fermentation of ethanol in the presence of bacteria. The equation representing this reaction is given below.
C2H5OH + O2 -> CH3COOH + H2O
In a small scale experiment using this second method it was found that 23g of ethanol produced only 4.54g of ethanoic acid. Calculate the percentage yield for this experiment."

The mark scheme then says: "1st mark - 23.0g of ethanol produces 30.0g ethanoic acid. 2nd mark - 15.1% (4.54 x 100 / 30)."
I'm fine with the second mark calculating yield, but how do you work out the theoretical production as 30g?? I don't get what I'm missing...

Solved: I was calculating the Mr for Ethanol wrong and it was screwing me up, but now I realise that you use the mass of Ethanol divided by the Mr to work out the moles, then times the moles by the Mr for ethanoic acid and it gives you the theoretical yield of 30g. Doy!

The second question is a simple ideal gas equation calculation. It states: "An excess pf sodium hydrogencarbonate was added to a 10.0cm3 sample of a 15.0 moldm-3 solution of ethanoic acid. Use the ideal gas equation and your answer from question 11 (a) to calculate the volume of carbon dioxide gas that would be formed at 20 degrees C and 1.00 x 10^5 Pa."
The question in 11(a) was to use the equation:
2CH3COOH + Na2CO3 -> 2CH3COONa + H2O +CO2
to calclulate moles of ethanoic acid in 10.0cm3 of 15.0 moldm-3 solution (which comes to 0.150mol).
Using the equation, I halved this amount to 0.075, as there are half the moles of Carbon Dioxide produced. However, in the mark scheme, it uses the value 0.150mol in the ideal gas equation, the same as that for ethanoic acid. What am I missing here?

Thanks for any help and I massively appreciate anything.
It's sodium hydrogen carbonate NOT sodium carbonate.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 14, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

### Who would you like to thank?

Poll
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.