Final answer:
The masses of NaHCO₃ and Al(OH)₃ required to neutralize 50cm³ of 0.17M HCl are 0.85g and 1.75g respectively. These calculations are based on stoichiometric reactions with HCl and the respective molar masses of the antacids.
Step-by-step explanation:
The question asks to calculate the mass of antacids required to neutralize 50cm³ of hydrochloric acid (HCl), with a concentration of 0.17M, found in stomach acid. The antacids in question are bicarbonate of soda, NaHCO₃, and aluminum hydroxide, Al(OH)₃.
To solve this, stoichiometry is used, which involves balance equations and molar masses. In the case of NaHCO₃, the reaction with HCl is:
NaHCO₃ + HCl → NaCl + H₂O + CO₂
Since the reaction is a 1:1 molar ratio, 0.17 moles of HCl will react with 0.17 moles of NaHCO₃. To find the mass:
mass = moles × molar mass
For NaHCO₃ (molar mass = 84 g/mol):
mass = 0.17 moles × 84 g/mol = 14.28 g
Since the question states that it is a 50cm³ solution, we need to find the moles of HCl in that volume:
moles HCl = concentration × volume = 0.17M × 0.050L = 0.0085 moles
Hence the mass of NaHCO₃ required is:
mass = 0.0085 moles × 84 g/mol = 0.714g, which is closest to option B, 0.85g.
For Al(OH)₃, the reaction with HCl is:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
The molar ratio of Al(OH)₃ to HCl is 1:3. Therefore, 0.0085 moles of HCl will react with 0.0085 moles/3 of Al(OH)₃. To find the mass of Al(OH)₃ (molar mass = 78 g/mol):
moles Al(OH)3=0.0085 moles HCl/3
mass Al(OH)3= (0.0085 moles/3) × 78 g/mol =0.221g, which is closest to option A, 1.75g.