A paracetamol tablet was crushed and an excess of HCl 25cm³ of 1.0mol was added. the result of the solution was titrate against 0.5mol of NaOH. calculate the mass of CaCO3 in the paracetamol.​

1.35g of CaCO₃ are in the paracetamol

Explanation:

Assuming the volume of the 0.5mol NaOH was 23cm³

The CaCO₃ of paracetamol reacts with HCl as follows:

CaCO₃ + 2HCl → CO₂ + H₂O + CaCl₂

When an excess of HCl is added and titrated, we can find the moles of HCl that react = Moles CaCO₃ based on the second reaction:

NaOH + HCl → H₂O + NaCl

Moles added of NaOH:

23cm³ = 0.023L * (0.5mol / L) = 0.0115 moles NaOH = Moles HCl in excess

Moles of HCl:

25cm³ = 0.025L * (1.0mol / L) = 0.025 moles HCl

The moles of HCl that reacted = Moles CaCO₃ are:

0.025 moles - 0.115 moles = 0.0135 moles CaCO₃

The mass of CaCO₃ -Molar mass: 100.09g/mol- is:

0.0135 moles CaCO₃ * (100.09g/mol) =

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