Question

Calculate the maximum volume in ml of 0.15M HCl that each of the following antacid formulations would be expected to neutralize. Assume complete neutralization. Show your working.

a. A tablet containing 150mg Mg(OH)2

b. A tablet containing 850mg of CaCO3

Answer 1

To calculate the maximum volume of 0.15M HCl that each antacid formulation can neutralize, use stoichiometry and the balanced chemical equations. a) A tablet containing 150mg Mg(OH)2 can neutralize 34.3 mL of HCl. b) A tablet containing 850mg CaCO3 can neutralize 113 mL of HCl.

Step-by-step explanation:

In order to calculate the maximum volume of 0.15M HCl that each antacid formulation can neutralize, we need to use stoichiometry and balanced chemical equations. Let's calculate the volume for each:

a. For a tablet containing 150mg Mg(OH)2:

First, convert the mass of Mg(OH)2 to moles:

Moles of Mg(OH)2 = mass / molar mass = 0.150g / 58.32g/mol = 0.00257 mol

Based on the balanced equation Mg(OH)2 + 2HCl → MgCl2 + 2H2O, we can see that 1 mole of Mg(OH)2 neutralizes 2 moles of HCl. So, the number of moles of HCl that can be neutralized is:

Moles of HCl = 2 * 0.00257 = 0.00514 mol

Finally, using the molarity of HCl, we can calculate the volume:

Volume of HCl = Moles of HCl / Molarity = 0.00514 mol / 0.15 mol/L = 34.3 mL

b. For a tablet containing 850mg of CaCO3:

First, convert the mass of CaCO3 to moles:

Moles of CaCO3 = mass / molar mass = 0.850g / 100.09g/mol = 0.008493 mol

Based on the balanced equation CaCO3 + 2HCl → CaCl2 + CO2 + H2O, we can see that 1 mole of CaCO3 neutralizes 2 moles of HCl. So, the number of moles of HCl that can be neutralized is:

Moles of HCl = 2 * 0.008493 = 0.01699 mol

Finally, using the molarity of HCl, we can calculate the volume:

Volume of HCl = Moles of HCl / Molarity = 0.01699 mol / 0.15 mol/L = 113 mL

Answer 2

a. 34 mL; b. 110 mL

a. A tablet containing 150 Mg(OH)₂

Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O

Moles of Mg(OH)₂ = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂

= 2.572 mmol Mg(OH)₂

Moles of HCl = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]

= 5.144 mmol HCl

Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl

b. A tablet containing 850 mg CaCO₃

CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O

Moles of CaCO₃ = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃

= 8.492 mmol CaCO₃

Moles of HCl = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]

= 16.98 mmol HCl

Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl