Question

An antacid tablet with an active ingredient of CaCO3 was dissolved in 50.0 mL of 0.300 M HCl. When this solution was titrated with 0.300 M sodium hydroxide, 32.47 mL was required to reach the end point. Determine the mass (in grams) of calcium carbonate in the tablet. Molar Mass of CaCO3

Answer 1

Answer:

Explanation:

mole of HCl remaining after reaction with CaCO₃

= .3 M of NaOH of 32.47 mL

= .3 x .03247 moles

= .009741 moles

Initial HCl taken = .3 x .005 moles = .0015 moles

Moles of HCl reacted with CaCO₃

= .009741 - .0015 = .008241 moles

CaCO₃     +    2HCl   =   CaCl₂  +  CO₂  +  H₂O .

1 mole        2 moles

2 moles of HCl reacts with 1 mole of  CaCO₃

.008241 moles of HCl reacts with .5 x .008241 moles of  CaCO₃

CaCO₃ reacted with HCl =  .5 x .008241 = .00412 moles

the mass (in grams) of calcium carbonate in the tablet

= .00412 x 100 = .412 grams . ( molar mass of calcium carbonate = 100 )