Gastric juice contains about 3.0 g HCI per litre. If a person produces about 2.5 L of gastric juice per day, how many antacid tablets each containing 400 mg of `Al (OH)_3` are needed to neutralise all the HCI produced in one day.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the total amount of HCl produced per day.
Given that gastric juice contains 3.0 g of HCl per liter and a person produces 2.5 L of gastric juice per day, we can calculate the total amount of HCl as follows:

\[ \text{Total HCl} = \text{Concentration of HCl} \times \text{Volume of gastric juice} \]
\[ \text{Total HCl} = 3.0 \, \text{g/L} \times 2.5 \, \text{L} = 7.5 \, \text{g} \]

### Step 2: Write the balanced chemical equation for the reaction between HCl and Al(OH)₃.
The balanced chemical equation is:

\[ 3 \, \text{HCl} + \text{Al(OH)}_3 \rightarrow \text{AlCl}_3 + 3 \, \text{H}_2\text{O} \]

This indicates that 3 moles of HCl react with 1 mole of Al(OH)₃.

### Step 3: Calculate the molar mass of HCl and Al(OH)₃.
- Molar mass of HCl:
- H: 1 g/mol
- Cl: 35.5 g/mol
- Total: 1 + 35.5 = 36.5 g/mol

- Molar mass of Al(OH)₃:
- Al: 27 g/mol
- O: 16 g/mol (3 O atoms)
- H: 1 g/mol (3 H atoms)
- Total: 27 + (3 × 16) + (3 × 1) = 27 + 48 + 3 = 78 g/mol

### Step 4: Calculate the amount of Al(OH)₃ needed to neutralize 7.5 g of HCl.
Using the stoichiometry from the balanced equation, we can set up the following proportion:

\[ \text{If } 3 \, \text{HCl} \text{ (36.5 g)} \text{ reacts with } 1 \, \text{Al(OH)}_3 \text{ (78 g)} \]
\[ \text{Then } 7.5 \, \text{g HCl reacts with } x \, \text{g Al(OH)}_3 \]

Using the cross-multiplication method:

\[ x = \frac{7.5 \, \text{g} \times 78 \, \text{g}}{3 \times 36.5 \, \text{g}} \]
\[ x = \frac{585}{109.5} \approx 5.34 \, \text{g} \]

### Step 5: Calculate the number of antacid tablets needed.
Each antacid tablet contains 400 mg of Al(OH)₃. First, convert this to grams:

\[ 400 \, \text{mg} = 0.4 \, \text{g} \]

Now, calculate the number of tablets required:

\[ \text{Number of tablets} = \frac{\text{Total Al(OH)}_3 \text{ needed}}{\text{Amount per tablet}} = \frac{5.34 \, \text{g}}{0.4 \, \text{g}} = 13.35 \]

Since we cannot have a fraction of a tablet, we round up to the nearest whole number:

\[ \text{Number of tablets needed} = 14 \]

### Final Answer:
**14 antacid tablets are needed to neutralize all the HCl produced in one day.**