Question

How many moles of stomach acid would be neutralized by one tablet of Rolaid that contains 550.mg of CaCO3 and 110.mg of mMg(OH)2 (molar mass = 58.3g/mol)

Answer 1

Balanced equations:

CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

Mg(OH)₂ + 2HCl --> MgCl₂ + 2H₂O

Moles of each component in a tablet:

$n(CaCO_3)=\frac{0.550g}{100.1g/mol}=0.00549mol$

$n(Mg(OH)_2)=\frac{0.110g}{58.3g/mol}=0.00189mol$

According to the equations, one mole of both CaCO₃ and Mg(OH)₂ neutralizes 2 moles of HCl (stomach acid). Therefore,

$n(HCl)=2\times0.00549mol+2\times0.00189mol=0.0148mol$

Answer: 0.0148 mol