CBSE Class 11-science Answered

Al(OH)₃ + 3HCl ? AlCl₃ + 3H₂O

1L gastric juice contains 3 grams HCl
2.5 litres gastric juice contains = 2.5 × 3 = 7.5g HCl

Molar mass of  HCl = 36.5g/mol
7.5g = 7.5/36.5 = 0.205 mol HCl
1mol Al(OH)₃ reacts with 3mol HCl
0.205 mol HCl reacts with 0.205/3 = 0.068 mol Al(OH)₃
0.068 mol Al(OH)₃ is required to neutralize all the HCl.
Molar mass Al(OH)₃ = 78g/mol
400mg = 0.400g Al(OH)₃
Each tablet contains 0.400/78 = 0.00513 mol Al(OH)₃
Number of tablets required = 0.068/0.00513 = 13.25 tablets.