A shell weighs 1.216g and 10cm3 of 5 mol dm-3 of HCl is added.after shell dissolved, solution is transferred and volume made up to 25cm3.

Question

The solution required 28 cm3 of 1 mol dm-3 sodium hydroxide for complete neutralisation.

I have calculated moles of moles of NaOH. I am having trouble working out how many moles of acid remained in the beaker after the reaction with the shell. 

 

The questions that is asked include how many moles of acid remained in the beaker after the reaction with the shell? I got 0.028 moles
The second question is how many moles of acid reacted with the shell? -----How would I work this out?

Answer 1

So is the question asking "how many moles of HCl remained", or is it asking something else?

If you know the moles of NaOH (which you said you calculated), then you can calculate the moles of HCl the same way.

moles HCl added to shell = 0.01 L x 5 mol/L = 0.05 moles HCl ADDED

HCl + NaOH ==> NaCl + H2O

Moles of NaOH = moles HCl so moles HCl remaining = initial moles HCl added - moles HCl reacting with NaOH.

 

Got it?  The rest of the HCl had reacted with the CaCO3 in the shell, and this is a BACK TITRATION.  You will ultimately probably be asked to calculate the amount (moles) of CaCO3 in the shell.

2HCl + CaCO3 ==> CaCl2 + CO2 + H2O

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