Question

An artificial Dolomite containing calcium carbonate and magnesium carbonate weighing 1.0200g was reacted with 50.0mL of 0.920M hydrochloric acid to produce a solution. The excess hydrochloric acid in the solution required 38.1mL of.0 630M sodium hydroxide to complete the titration. Calculate the percentage of each component in the mixture.

Answer 1

No of mole of HCl reacted = (50*0.92/1000)-(38.1*0.63/1000) = 0.022 mol

so that,

from balanced equation , 1 mol CaCo3 = 2 mol HCl

                                         1 mol MgCO3 = 2 mol HCl

No of mole of HCl reacted = 2*No of mol of CaCO3 + 2*No of mol of MgCO3

          0.022 = (2*(x/100.1))+(2*((1.02-x)/84.314))

x =0.587

mass of CaCO3 in sample = 0.587 g

mass of MgCO3 in sample = 1.02-0.587 = 0.433 g

%of MgCO3 = 0.433/1.02*100 = 42.45%

% of CaCO3 = 0.587/1.02*100 = 57.55%