Some over-the-counter antacids contain aluminum hydroxide, whereas others contain calcium carbonate. If a tablet contains 750 mg of Al (OH) 3, and another contains 750 mg of CaCO3, which consumes the greater mass of HCl?

AL (OH) 3 consumes greter mass of HCl.

Explanation:

Equation of the reaction

CaCO3 (s) + HCl (aq) - -> CaCl2 (aq) + CO2 (g) + H2O (g)

Mass of CaCO3 = 750 mg

= 0.75 g

Molar mass = 40 + 12 + (16*3)

= 100 g/mol.

Since 1 mole of CaCO3 reactes with 1 mole of HCl to form 1 mole of CaCl2.

Number of moles = mass/molar mass

= 0.75/100

= 0.0075 moles of CaCO3

= 0.0075 moles of HCl

Molar mass of HCl = 1 + 35.5

= 36.5 g/mol

Mass of HCl consumed = moles*molar mass

= 0.0075*36.5

= 0.27 g of HCl

Equation of reaction

Al (OH) 3 (aq) + 3HCl (aq) - -> AlCl3 (aq) + 3H2O (g)

Mass of Al (OH) 3 = 750 mg

= 0.75 g

Molar mass = 27 + (1 + 16) * 3

= 78 g/mol.

Since 1 mole of Al (OH) 3 reactes with 3 mole of HCl to form 1 mole of AlCl3.

Number of moles = mass/molar mass

= 0.75/78

= 0.0096 moles of Al (OH) 3

= 0.0096*3

= 0.02885 moles of HCl

Molar mass of HCl = 1 + 35.5

= 36.5 g/mol

Mass of HCl consumed = moles*molar mass

= 0.02885*36.5

= 1.05 g of HCl

Therefore, Al (OH) 3 consumes greater mass of HCl.

750 mg of Al (OH) 3 consumes greater mass of HCl

Explanation:

Equation of reaction between CaCO3 and HCl:

CaCO3 + 2HCl = CaCl2 + CO2 + H2O

1 mole (100g) of CaCO3 consumes 2 moles (73g) of HCl

750mg (0.75g) of CaCO3 will consume 0.75*73/100 = 0.5475g of HCl

Equation of reaction between Al (OH) 3 and HCl:

Al (OH) 3 + 3HCl = AlCl3 + 3H2O

1 mole (78g) of Al (OH) 3 consumes 3 mole (109.5g) of HCl

750mg (0.75g) of CaCO3 will consume 0.75*109.5/78 = 1.0529g of HCl

750mg of CaCO3 consumes 0.5475g of HCl while 750mg of Al (OH) 3 consumes 1.0529g of HCl