Question

Mg(OH)\(_2\) is used as an antaci(D) If a person, suffering from acidity, produces 2.5 L of Gastric juice in a day, approximately how many antacid tablets, each containing 600 mg of Mg(OH)\(_2\), will be required to completely neutralize the whole HCl produced in the stomach in one day? (Gastric juice contains 3.0 g of HCl per L). (Atomic masses: Mg = 24 g/mol, O = 16 g/mol, and H = 1 g/mol.)

• A. 10 tablets
• B. 4 tablets
• C. 7 tablets
• D. 6 tablets

Hint:

When dealing with neutralization reactions, always balance the equation to relate the amount of acid and base, and use molar mass to convert between grams and moles.

Answer 1

The Correct Option is A

1. Moles of HCl produced in 1 day:
Gastric juice contains 3.0 g of HCl per L. So, in 2.5 L of gastric juice, the total amount of HCl is: \[ \text{Amount of HCl} = 3.0 \, \text{g/L} \times 2.5 \, \text{L} = 7.5 \, \text{g} \] Now, the molar mass of HCl is \(1 + 35.5 = 36.5 \, \text{g/mol}\), so the moles of HCl are: \[ \text{Moles of HCl} = \frac{\text{Mass of HCl}}{\text{Molar Mass of HCl}} = \frac{7.5 \, \text{g}}{36.5 \, \text{g/mol}} = 0.205 \, \text{mol} \] 2. Moles of Mg(OH)\(_2\) required to neutralize the HCl:
The balanced equation for the neutralization reaction is: \[ \text{Mg(OH)}_2 + 2 \text{HCl} \rightarrow \text{MgCl}_2 + 2 \text{H}_2\text{O} \] This shows that 1 mole of Mg(OH)\(_2\) reacts with 2 moles of HCl. Therefore, the moles of Mg(OH)\(_2\) required are: \[ \text{Moles of Mg(OH)}_2 = \frac{0.205 \, \text{mol HCl}}{2} = 0.1025 \, \text{mol Mg(OH)}_2 \] 3. Mass of Mg(OH)\(_2\) required:
The molar mass of Mg(OH)\(_2\) is \(24 + 2(16 + 1) = 58 \, \text{g/mol}\), so the mass of Mg(OH)\(_2\) required is: \[ \text{Mass of Mg(OH)}_2 = 0.1025 \, \text{mol} \times 58 \, \text{g/mol} = 5.95 \, \text{g} \] 4. Number of tablets required:
Each tablet contains 600 mg (or 0.6 g) of Mg(OH)\(_2\). Therefore, the number of tablets required is: \[ \text{Number of tablets} = \frac{\text{Mass of Mg(OH)}_2}{\text{Mass per tablet}} = \frac{5.95 \, \text{g}}{0.6 \, \text{g}} = 9.92 \approx 10 \, \text{tablets} \] Thus, approximately 10 tablets are required to neutralize the whole HCl produce(D)