Answer:
[tex]\boxed{\text{0.20 g}}[/tex]
Explanation:
We know we will need a balanced chemical equation with molar masses, volumes, and concentrations, so, let's gather all the information in one place.
M_r: 84.01
HCl + NaHCO₃ ⟶ NaCl + H₂O + CO₂
V/mL: 200.
c/mol·L⁻¹: 0.012
(a) Moles of HCl
[tex]\text{Moles of HCl} =\text{0.200 L HCl} \times \dfrac{\text{0.012 mol HCl}}{\text{1 L HCl}}\\\\=\text{0.0024 mol HCl}[/tex]
(b) Moles of NaHCO₃
The molar ratio is 1 mol NaHCO₃ = 1 mol HCl
[tex]\text{Moles of NaHCO$_{3}$}= \text{0.0024 mol HCl} \times \dfrac{\text{1 mol {NaHCO$_{3}$}}}{ \text{1 mol HCl}}\\\\= \text{0.0024 mol NaHCO$_{3}$}[/tex]
(c) Mass of NaHCO₃
[tex]\text{Mass of NaHCO$_{3}$}= \text{0.0024 mol NaHCO$_{3}$} \times \dfrac{\text{84.01 g {NaHCO$_{3}$}}}{ \text{1 mol NaHCO$_{3}$}}\\\\= \textbf{0.20 g NaHCO$_{3}$}\\\\\text{The man would need to ingest }\boxed{\textbf{0.20 g}} \text{ of NaHCO$_{3}$}.[/tex]
