Question

2. Al(OH),(s) + 3 HCl(aq) à 3 H2O(l) + AlCl3(aq). This reaction shows how aluminum hydroxide

Answer 1

This tablet insufficient to neutralize the acid in the patient's stomach

Further explanation

Reaction

Al(OH)₃(s) + 3HCl(aq) ⇒ 3 H₂O(l) + AlCl₃(aq)

0.25 g of  Aluminum hydroxide-Al(OH)₃ , mol (MW=78 g/mol) :

$\tt \dfrac{0.25}{78}=0.00321$

mol HCl : mol Al(OH)₃ = 3 : 1

$\tt mol~HCl=3\times 0.00321=0.00963$

mass HCl (MW=36,46 g/mol) :

$\tt 0.00963\times 36.46=0.351$

0.351 < 0.88⇒this tablet insufficient to neutralize the acid