Titrations: magnesium carbonate and hydrochloric acid

I have been given a homework question and have found answers to all part but the last one.. can you assess my answers and offer advice as to how to do the final part.

The equation for the reaction between magnesium carbonate and hydrochloric acid is given below.

MgCO3 + 2HCl → MgCl2 + H2O + CO2

When 75.0cm3 of 0.500mol.dm-3 hydrochloric acid were added to 1.25g of impure MgCO3 some acid was left unreacted. This unreacted acid required 21.6cm3 of 0.500mol.dm-3 solution of sodium hydroxide for complete reaction.

1. Calculate the number of moles of HCl in 75.0cm3 of 0.500mol.dm-3 hydrochloric acid.
2. Calculate the number of moles of NaOH used to neutralise the unreacted HCl.
3. Show that the number of moles of HCl which reacted with the MgCO3 in the sample was 0.0267
4. Calculate the number of moles and the mass of MgCO3 in the sample, and hence deduce the percentage by mass of MgCO3 in the sample.

I) cxv= number of moles
(75cm/1000=0.075dm)
0.500x0.075dm=0.0375moles
ii) cxv= number of moles
(21.6cm/1000=0.0216)
0.500x0.0216dm=0.0108
iii) 0.0375moles in original HCL
0.0108 moles of NaOH was needed to neutralise unreacted HCl: 0.0108 moles of
HCl was unreacted so
0.0375-0.0108= 0.0267 moles reacted with the MgCO3
iv) THIS IS THE PART I HAVE DIFFICULTY WITH
when it says "in the sample" what part is it referring to because its asking for the
mass of MgCO3 which is given in the question "1.25g". Im quite confused.

If anyway could help explain who you go about answering this question (and other similar ones) id greatly appreciate it.

 joinforfun8909 Posts: 12, Reputation: 10 New Member #2 Apr 15, 2010, 11:11 AM

1.25 g is not mass of MgCO3. It is the mass of MgCO3 with impurities. So the task is to find the mass of MgCO3 actually present.

That can be done by using the number of moles of HCl consumed [assuming that the impurities don't react with HCl! ]. You know that 1 mole of HCl reacts with 2 moles of MgCO3. So from that you can get the answer

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