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# How do I work out the moles?

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1. Hi!

I'm trying to do a homework and I don't know how to work out the missing moles of this reactant (see picture Q7). Could someone please help?

Blake

Information I have which might help...

Mass of 5cm^3
HCl = 5.25g
Water = 5g
Ethyl ethanoate = 4.55g
Ethanol = 3.98g
Glacial Ethanoic acid = 5.2g

Titre of NaOH = 42.2cm^3

Titre of HCl = 14.3cm^3

Moles of ethyl ethanoate = 0.052

Mass of water in HCl = 4.72805g

Moles of H2O from HCl = 0.262669

Amount of NaOH needed to neutralise ethanoic acid in equilibrium mixture = 27.9cm^3

Moles of ethanoic acid produced = 0.0279

Started with 0 moles of ethanol and 0 moles of ethanoic acid

Starting mixture contained 5cm^3 of HCl and 5cm^3 of ethyl ethanoate
2. Ah it's an equilibrium question. Well done, you've done the hard bit of calculating how much ethanoic acid is produced! Now think- if you started with 0 moles of both ethanol and ethanoic acid and at equilibrium the amount of ethanol has increased to 0.0279 mole, what is the number of moles of ethanoic acid? Look at their molar ratio- in a reaction you can't just produce one of the products/reactants, you produce all of them
3. (Original post by richpanda)
Ah it's an equilibrium question. Well done, you've done the hard bit of calculating how much ethanoic acid is produced! Now think- if you started with 0 moles of both ethanol and ethanoic acid and at equilibrium the amount of ethanol has increased to 0.0279 mole, what is the number of moles of ethanoic acid? Look at their molar ratio- in a reaction you can't just produce one of the products/reactants, you produce all of them
Oh! So that's 0.0279 mol too?
4. (Original post by Blake Jones)
Oh! So that's 0.0279 mol too?
yes

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