Stomach acid is approximately \(0.020 \mathrm{M} \mathrm{HCl}\). What volume of this acid is neutralized by an antacid tablet that weighs \(330 \mathrm{mg}\) and contains \(41.0 \% \mathrm{Mg}(\mathrm{OH})_{2}, 36.2 \% \mathrm{NaHCO}_{3}\), and \(22.8 \% \mathrm{NaCl} ?\) The reactions in- volved are $$ \begin{gathered} \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Mg}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O} \\ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \end{gathered} $$

First, we need to find out how much of each active ingredient is present in the antacid tablet. Given that the tablet weighs \(330\,\text{mg}\) and contains \(41.0 \%\) \(Mg(OH)_2\) and \(36.2 \%\) \(NaHCO_3\): $$\text{Mass of}\, Mg(OH)_2 = 0.410 \times 330\,\text{mg} = 135.3\,\text{mg}$$ $$\text{Mass of}\, NaHCO_3 = 0.362 \times 330\,\text{mg} = 119.46\,\text{mg}$$ Now we can convert these masses into moles by using their molar masses: \(Mg(OH)_2 = 58.32 \,g/mol\), \(NaHCO_3 = 84.01\, g/mol\) $$\text{Moles of}\, Mg(OH)_2 = \frac{135.3\,\text{mg}}{58.32\, g/mol} \times \frac{1\,g}{1000\,\text{mg}} = 0.00232\,\text{mol}$$ $$\text{Moles of}\, NaHCO_3 = \frac{119.46\,\text{mg}}{84.01\, g/mol} \times \frac{1\,g}{1000\,\text{mg}} = 0.00142\,\text{mol}$$

We can use the given chemical reactions and stoichiometry to find out how many moles of \(H^+\) ions are neutralized by \(Mg(OH)_2\) and \(NaHCO_3\). From the reactions: $$Mg(OH)_2 + 2H^+ \rightarrow Mg^{2+} + 2H_2O$$ $$HCO_3^- + H^+ \rightarrow CO_2 + H_2O$$ There is a \(1:2\) ratio between moles of \(Mg(OH)_2\) and moles of \(H^+\) ions, and a \(1:1\) ratio between moles of \(HCO_3^-\) and moles of \(H^+\) ions. $$\text{Moles of}\, H^+\, \text{neutralized by}\, Mg(OH)_2 = 2 \times 0.00232\,\text{mol} = 0.00464\,\text{mol}$$ $$\text{Moles of}\, H^+\, \text{neutralized by}\, NaHCO_3 = 1 \times 0.00142\,\text{mol} = 0.00142\,\text{mol}$$ Total moles of \(H^+\) ions neutralized by the antacid tablet: $$\text{Total moles of}\, H^+= 0.00464\,\text{mol} + 0.00142\,\text{mol} = 0.00606\,\text{mol}$$

Now, we can apply the formula \(C = \frac{n}{V}\), where \(C\) is the concentration of \(HCl\) in stomach acid and \(n\) is the moles of \(H^+\) ions neutralized. Rearranging the formula to find \(V\): $$V = \frac{n}{C}$$ Substituting the values, \(C = 0.020\,\text{M}\) and \(n = 0.00606\,\text{mol}\): $$V = \frac{0.00606\,\text{mol}}{0.020\,\text{M}} = 0.303\,\text{L}$$ This means that the volume of stomach acid neutralized by the antacid tablet is approximately \(0.303\,\text{L}\).