Answer:
[tex]3.8x10^{-1}gNaHCO_3[/tex]
Explanation:
Hello,
This is a typical case in which the following chemical reaction is carried out:
[tex]HCl+NaHCO_3-->NaCl+H_2O+CO2[/tex]
Since 50 mL of a 0.091M solution of HCl is employed, we perform the shown below stoichiometric calculation to find the sodium hydrogen carbonate grams that must be ingested by the woman:
[tex]m_{NaHCO_3}=0.050L*0.091\frac{molHCl}{L}*\frac{1molNaHCO_3}{1molHCl}*\frac{84gNaHCO_3}{1molNaHCO_3} =3.8x10^{-1}gNaHCO_3[/tex]
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