You added 75cm^3 of 0.5moldm^-3 HCl to impure MgCO2, and some was left unreacted. The unreacted HCl reacted completely with 21.6cm^m of 0.5moldm^-3 NaOH. So what is the percentage purity of the MgCO3 sample?

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 moldm^3-3 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%