Question

2. The approximate concentration of hydrochloric acid, HCl, in the stomach (stomach

Answer 1

Answer:

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) =

0.221g of Al(OH)₃ are required