2. The approximate concentration of hydrochloric acid, HCl, in the stomach (stomach acid) is 0.17M. Calculate the mass of the following antacids required to neutralize 50cm of
stomach acid.
(a) Bicarbonate of soda NaHCO3
(b) Aluminum hydroxide, Al(OH)3

Please .​

a) 0.714g of bicarbonate of soda are required.

b) 0.221g of Al(OH)₃ are required

Explanation:

The reactions of HCl with bicarbonate of soda and aluminium hydroxide are:

HCl + NaHCO₃ → H₂O + NaCl + CO₂

3 HCl + Al(OH)₃ → 3H₂O + AlCl₃

The moles of HCl that we need neutralize are:

50mL = 0.050L * (0.17mol / L) = 0.0085 moles HCl

To solve these problem we need to find the moles of the antacid using the chemical reaction and its mass using its molar mass;

a) Moles NaHCO₃ = Moles HCl = 0.0085 moles

The mass is -Molar mass NaHCO₃: -84g/mol-

0.0085 moles * (84g / mol) = 0.714g of bicarbonate of soda are required

b) 0.0085 moles HCl * (1mol Al(OH)₃ / 3mol HCl) = 2.83x10⁻³ moles Al(OH)₃

The mass is -Molar mass: 78g/mol-:

2.83x10⁻³ moles Al(OH)₃ * (78g/mol) =

20 drops hope this

10 years is the answer