Imagine if the HCl used in this experiment was prepared at a concentration of 0.150M. Using your calculated volume of HCl neutralized by your whole tablet, calculate the mass of active ingredient (CaCO3) in the tablet in milligrams. Show your work to receive credit.

Explanation:

Given:

Concentration of HCl = 0.15 M

Volume of HCl neutralising the whole tablet, Va = 47.7 mL

Equation of the reaction:

CaCO3 + 2HCl --> CaCl2 + H2O + CO2

Number of moles = concentration × volume

= 0.15 × 47.7 × 10^-3

= 0.007155 mol

By stoichiometry, 1 mole of CaCO3 reacted completely with 2 moles of HCl. Therefore, number of moles of CaCO3 that occurred in the neutralisation reaction = 0.007155 mole of HCl × 1 moles of CaCO3/2 mole of HCl

= 0.0035775 moles of CaCO3

Molar mass of CaCO3 = 40 + 12 + (3 × 16)

= 100 g/mol.

Mass = molar mass × number of moles

= 100 × 0.0035775

= 0.35775 g

= 0.36 g of CaCO3.

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