2.13 g of a carbonate M2CO3 was dissolved in water and made up to 100 cm3
a volumetric flask. 10.0 cm3 of this solution required 18.50 cm3 of 0.100 mol dm−3
hydrochloric acid for neutralisation.
The equation below represents the reaction between the acid and the carbonate:
2HCl(aq) M2CO3(aq) → 2MCl(aq) H2O(l) CO2(g)
a Calculate the amount in moles of HCl needed to neutralise 10.0 cm3 of the
b Using the equation above, write down the amount in moles of M2CO3 used
in the titration.
c Calculate the amount in moles of M2CO3 in 100 cm3 of the carbonate
d Calculate the molar mass of M2CO3
e Calculate the relative atomic mass of M. Give your answer to one decimal
f Use the Periodic Table to identify M in the formula M2CO3
HOW DO I DO C?