Answer:
1.35g of CaCO₃ are in the paracetamol
Step-by-step 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) =
1.35g of CaCO₃ are in the paracetamol