To determine the moles of [tex]\( CO_2 \)[/tex] produced during the reaction, we'll start with the balanced chemical equation provided:
[tex]\[ 2 \text{HCl} \, (aq) + \text{Na}_2\text{CO}_3 \, (aq) \rightarrow 2 \text{NaCl} \, (aq) + \text{H}_2\text{O} \, (l) + \text{CO}_2 \, (g) \][/tex]
From the balanced equation, it is clear that 2 moles of HCl react to produce 1 mole of [tex]\( CO_2 \)[/tex].
In the previous step, we determined that 0.10 moles of HCl are reacting.
Since the stoichiometry of the reaction shows that 2 moles of HCl produce 1 mole of [tex]\( CO_2 \)[/tex], we can calculate the moles of [tex]\( CO_2 \)[/tex] produced by taking half of the moles of HCl reacting:
[tex]\[ \text{Moles of } CO_2 = \frac{\text{moles of } HCl}{2} \][/tex]
Substituting the given value:
[tex]\[ \text{Moles of } CO_2 = \frac{0.10 \, \text{moles of } HCl}{2} = 0.05 \, \text{moles} \][/tex]
Thus, the moles of [tex]\( CO_2 \)[/tex] produced during the reaction are:
[tex]\[ \boxed{0.05} \, \text{moles} \][/tex]
This completes our determination of the moles of [tex]\( CO_2 \)[/tex] generated from the reaction.
