2.75 g sample of dolomite containing CaCO3 and MgCO3 is dissolved in 80 cm3 of 1M HCl solution. The solution is then diluted to 250 cm3. 25 cm3 of this solution requires 20 cm3 of 0.1M NaOH solution for complete neutralisation. Calculate the % composition of the sample.Please show how its worked out
NaOH + HCl ------ NaCl + H2O Number of moles for HCl 1M * 0.08 dm3 (divide by 1000 for the volume as you work with dm3/L not mL/cm3 in the process of titration) 0.08 mol in original sample NaOH is the limiting reagent and HCl is in excess. Number of moles for NaOH in diluted sample 0.1 M * 0.02 dm3 (divide by 1000 for the volume as you work with dm3/L not mL/cm3 in the process of titration) 0.002 mol in diluted sample. Ratio of HCl and NaOH is 1:1 so number of moles for HCl is the same, 0.002 mol in diluted sample. To reach the original sample, we multiply 0.002 mol by (250/25) 0.02 mol of HCl in original sample Amount of HCl reacted with CaCO3 and MgCO3 Excess - amount reacted with NaOH 0.08 mol - 0.02 mol 0.06 moles reacted with CaCO3 and MgCO3 Refer to balanced equation: MgCO3 + CaCO3 + 4HCl -------------- MgCl2 + CaCl2 + 2H2O + 2CO2 Ratio of MgCO3 CaCO3 between HCl is 1:4, this means that if we want the number of moles for MgCO3 or CaCO3 we have to divide by 4 since there is 4 HCl's and 1 of each CaCO3 and MgCO3 so MgCO3 0.06 mol / 4 0.015 moles and CaCO3 being the same 0.06 mol / 4 0.015 moles mass number of moles * molar mass mass for MgCO3 0.015 moles * 84.3139 g/mol 1.26 grams (3sf) mass for CaCO3 0.015 moles * 100.0869 g/mol 1.50 grams (3sf) % yield for MgCO3 1.26 g / 2.75 g * 100 45.8 % % yield for CaCO3 1.50 g / 2.75 g * 100 54.5 % Hope this helps!! :)
