Question

what is the net ionic equation for the reaction below?
pb(no₃)₂(aq) + na₂cro₄(aq) → pbcro₄(s) + 2nano₃(aq)
2na⁺(aq) + cro₄²⁻(aq) → na₂cro₄(s)
na⁺(aq) + no₃⁻(aq) → nano₃(aq)
pb(no₃)₂(aq) + na₂cro₄(aq) → pbcro₄(s) + 2nano₃(aq)
pb²⁺(aq) + cro₄²⁻(aq) → pbcro₄(s)
pb²⁺(aq) + 2 no₃⁻(aq) → pb(no₃)₂(aq)

Explanation:

Step1: Write the total ionic equation

$$\mathrm{Pb^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+CrO_4^{2 - }(aq)\to PbCrO_4(s)+2Na^+(aq)+2NO_3^-(aq)}$$

Step2: Cancel out the spectator ions ($\mathrm{Na^+}$ and $\mathrm{NO_3^-}$)

$$\mathrm{Pb^{2+}(aq)+CrO_4^{2 - }(aq)\to PbCrO_4(s)}$$

Answer:

$\mathrm{Pb^{2+}(aq)+CrO_4^{2 - }(aq)\to PbCrO_4(s)}$