Question

question 5 (2 points)
what is the net ionic equation for the reaction of sodium carbonate with iron (ii) nitrate?
fe²⁺ (aq) + co₃²⁻ (aq) → feco₃ (s)
na⁺ (aq) + no₃⁻ (aq) → nano₃ (s)
fe³⁺ (aq) + 3co₃⁻ (aq) → fe(co₃)₃ (s)
fe²⁺ (aq) + 2co₃⁻ (aq) → fe(co₃)₂ (s)

Explanation:

Step1: Write the balanced molecular equation

Sodium carbonate ($Na_2CO_3$) reacts with iron (II) nitrate ($Fe(NO_3)_2$) to form sodium nitrate ($NaNO_3$) and iron (II) carbonate ($FeCO_3$). The molecular equation is $Na_2CO_3(aq)+Fe(NO_3)_2(aq)
ightarrow 2NaNO_3(aq)+FeCO_3(s)$.

Step2: Write the complete ionic equation

Break all aqueous - soluble compounds into ions. $2Na^+(aq)+CO_3^{2 - }(aq)+Fe^{2+}(aq)+2NO_3^-(aq)
ightarrow 2Na^+(aq)+2NO_3^-(aq)+FeCO_3(s)$.

Step3: Write the net ionic equation

Cancel out the spectator ions ($Na^+$ and $NO_3^-$). The net ionic equation is $Fe^{2+}(aq)+CO_3^{2 - }(aq)
ightarrow FeCO_3(s)$.

Answer:

$Fe^{2+}(aq)+CO_3^{2 - }(aq)
ightarrow FeCO_3(s)$