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We wish to determine the moles of [tex]CO_2[/tex] produced when 50.0 mL of 2.0 M HCl reacts with excess [tex]Na_2CO_3[/tex].

[tex]\[
\begin{aligned}
2 HCl (aq) + Na_2CO_3 (aq) \rightarrow 2 NaCl (aq) + H_2O (l) + CO_2 (g)
\end{aligned}
\][/tex]

In the previous step, you determined 0.10 mol of HCl react.

How many moles of [tex]CO_2[/tex] form during the reaction?

Moles of [tex]CO_2[/tex]: ________

Sagot :

GinnyAnsweridn
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.
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