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2017 Question 3bii

Hi there, could you please show the working for this question I am struggling to understand it.

My working so far is:

n (CH3COOH) = cv = 0.166 x 0.02 = 3.32 x 10^-3 mol
c (CH3COOH) = mol/total volume = 3.32 x 10^-3 mol/0.03 = 0.1107 mol l^-1

Thank you :slight_smile:

Kia ora Shank21,

Have you had a look at the schedule already?

Is there anything is the working there that you would like me to explain?

Hi there,

Thanks, I’ve figured it out. I realised that n (NaOH) = n (CH3COO-) as NaOH is the limiting reagent.

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Well done! Fire away if you have any further questions.

Thanks! Just a general question:
If you have 1HCOOH + 3NaOH β†’ HCOONa + H2O. At the end of the reaction do you have 2 moles of NaOH remaining along with 1 mole of HCOONa and 1 mole of H20?

Yes, that looks correct. Here’s how I would lay out my working:

Thank you really appreciate it

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