Question

question 1
what are the spectator ions in the reaction of na₂co₃ and hcl in water?
□ h+
□ cl⁻
□ na⁺
□ hco₃⁻
□ co₃²⁻

Explanation:

Step1: Write the reaction equation

First, write the balanced chemical equation for the reaction between $\ce{Na_{2}CO_{3}}$ and $\ce{HCl}$ in water. The reaction is: $\ce{Na_{2}CO_{3}(aq) + 2HCl(aq) -> 2NaCl(aq) + H_{2}O(l) + CO_{2}(g)}$.

Step2: Write the ionic equation

Next, write the complete ionic equation by dissociating all strong electrolytes into their ions. $\ce{Na_{2}CO_{3}}$ dissociates into $\ce{2Na^{+}(aq) + CO_{3}^{2-}(aq)}$, and $\ce{HCl}$ dissociates into $\ce{2H^{+}(aq) + 2Cl^{-}(aq)}$. The products: $\ce{NaCl}$ dissociates into $\ce{2Na^{+}(aq) + 2Cl^{-}(aq)}$, $\ce{H_{2}O}$ is a liquid, and $\ce{CO_{2}}$ is a gas. So the complete ionic equation is: $\ce{2Na^{+}(aq) + CO_{3}^{2-}(aq) + 2H^{+}(aq) + 2Cl^{-}(aq) -> 2Na^{+}(aq) + 2Cl^{-}(aq) + H_{2}O(l) + CO_{2}(g)}$.

Step3: Identify spectator ions

Spectator ions are ions that appear on both sides of the complete ionic equation and do not participate in the chemical reaction (they do not change in state or form). Looking at the ionic equation, $\ce{Na^{+}}$ and $\ce{Cl^{-}}$ are present on both the reactant and product sides. $\ce{H^{+}}$ reacts with $\ce{CO_{3}^{2-}}$ to form $\ce{H_{2}O}$ and $\ce{CO_{2}}$, so they are not spectator ions. $\ce{CO_{3}^{2-}}$ reacts, and $\ce{HCO_{3}^{-}}$ is not a product here (the reaction goes to $\ce{CO_{2}}$ and $\ce{H_{2}O}$ from $\ce{CO_{3}^{2-}}$ and $\ce{H^{+}}$ in excess, but even in the first step, $\ce{CO_{3}^{2-}}$ reacts). So the spectator ions are $\ce{Na^{+}}$ and $\ce{Cl^{-}}$.

Answer:

The spectator ions are $\ce{Na^{+}}$ (sodium ion) and $\ce{Cl^{-}}$ (chloride ion). So the correct options are:
$\ce{Na^{+}}$ (the option with $\ce{Na^{+}}$) and $\ce{Cl^{-}}$ (the option with $\ce{Cl^{-}}$)