Answer to Question #161827 in General Chemistry for self

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