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;
<em>a) </em><em>Moles NaHCO₃ = Moles HCl = 0.0085 moles </em>
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) =
<h3>0.221g of Al(OH)₃ are required</h3>
