Answer:
Mass of NaHCO₃ required for the neutralization = 0.143 g
Step-by-step explanation:
The reaction between HCl and NaHCO₃ is given as
HCl + NaHCO₃ → NaCl + H₂O + CO₂
The fluid in the stomach of a man suffering from indigestion is considered to be 50 mL of a 0.034 M HCl solution. The mass of NaHCO₃ he would need to ingest to neutralize this much HCl.
We forst need to calculate the number of moles of HCl in 50.0 mL of 0.034 M HCl
(Number of moles) = (Conc in mol/L) × (Volume in L)
Conc in mol/L = 0.034 M
Volume in L = (50/1000) = 0.05 L
Number of moles of HCl = 0.034 × 0.05 = 0.0017 moles
From the stoichiometric balance of the reaction,
1 mole of HCl requires 1 mole of NaHCO₃
0.0017 moles of HCl will require 0.0017 moles of NaHCO₃.
So, we can then.calculate the mass of NaHCO₃ required for this neutralization.
Mass = (Number of moles) × (Molar mass)
Molar mass of NaHCO₃ = 84.007 g/mol
Mass of NaHCO₃ required for this neutralization = 0.0017 × 84.007 = 0.1428119 g = 0.143 g to 3 s.f because calculated values usually have 1 more significant figure than the given parameters for the calculations.
Hope this Helps!!!