MgCO3 +2HCl --> MgCl2 + H2O +CO2
For this kind of question, I would always try to convert what I can into moles because that is the link between the solid (the grams of the MgCO3) and the liquid (the HCL and NaOH). Here is my process:
Since mol=conc x vol, the mol of HCl initially added to the impure MgCO3 is 0.5moldm-3 x 75x10-3 dm3 = 0.0375 moles.
The mol of NaOH that reacted with the excess HCl moldm3-3 is 0.5 x 21.6x10-3 dm3 = 0.0108 moles.
Because the NaOH and HCL are in a 1 to 1 molar ratio: (1NaOH + 1HCl --> NaCl +H2O), the mole of excess HCl is also 0.0108.
(All in moles) Total HCl initially added to MgCO3 sample – HCl that actually reacted with (the pure) MgCO3 = Excess HCl (which was reacted with the NaOH).
So, 0.0375 - HCl that actually reacted with (the pure) MgCO3 = 0.0108.
HENCE, HCl that actually reacted with (the pure) MgCO3 = 0.0375 – 0.0108 = 0.0267 moles.
As the (pure)MgCO3 and HCl reacted in a 1:2 ratio (see equation above) the mol of MgCO3 is 0.0267/2 = 0.01335
The molecular mass (Mr) of (pure) MgCO3 is 84.3 (worked out from the periodic table)
So as the mass(of pure MgCO3) = mol x Mr, the mass = 0.01335 mol x 84.3 = 1.125g
Since the mass of impure MgCO3 is 1.25g, the percentage mass of pure MgCO3 is 1.125/1.25 x100% = 90%