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
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, Iβve 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. Hereβs how I would lay out my working:
Thank you really appreciate it
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