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

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.

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