# 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?

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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) MgCO= 0.0108.
HENCE, HCl that actually reacted with (the pure) MgCO3 = 0.0375 – 0.0108 = 0.0267 moles.

As the (pure)MgCOand 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 MgCOis 1.25g, the percentage mass of pure MgCOis 1.125/1.25 x100% = 90%

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