U4.2 · Net Ionic EquationsPractice · 净离子方程式

Soluble salts are fully dissociated. $\ce{K+}$ and $\ce{NO3-}$ are in solution before and after, so they are spectator ions and drop out, leaving $\ce{Ba^2+(aq) +}$ $\ce{CO3^2-(aq) ->}$ $\ce{BaCO3(s)}$. Choice A keeps the spectators; B is wrong because $\ce{KNO3}$ is soluble and doesn’t precipitate; D has the wrong formula and unequal charges.

HCl is a strong acid, fully ionized into $\ce{H+}$ and $\ce{Cl-}$; $\ce{NH3}$ is a weak base that stays mostly as molecules, so it is written as $\ce{NH3}$. $\ce{Cl-}$ is a spectator. The answer is $\ce{NH3(aq) +}$ $\ce{H+(aq) ->}$ $\ce{NH4+(aq)}$. B treats ammonia as a strong base full of $\ce{OH-}$; C writes the strong acid HCl as molecules; D keeps the spectator $\ce{Cl-}$.

The complete ionic equation is $\ce{Mg^2+(aq) +}$ $\ce{2Cl-(aq) +}$ $\ce{2Na+(aq) +}$ $\ce{2OH-(aq) ->}$ $\ce{Mg(OH)2(s) +}$ $\ce{2Na+(aq) +}$ $\ce{2Cl-(aq)}$. $\ce{Na+}$ and $\ce{Cl-}$ appear unchanged on both sides, so they are spectators; $\ce{Mg^2+}$ and $\ce{OH-}$ combine into the precipitate. Removing the spectators gives $\ce{Mg^2+(aq) +}$ $\ce{2OH-(aq) ->}$ $\ce{Mg(OH)2(s)}$.

Each Al loses 3 electrons to become $\ce{Al^3+}$, and each $\ce{Cu^2+}$ gains 2 to become Cu; electrons lost must equal electrons gained: 2 Al lose 6 and 3 $\ce{Cu^2+}$ gain 6. A conserves atoms, but its charge is +2 on the left and +3 on the right. C is a molecular equation, not a net ionic one, and D keeps the spectator $\ce{SO4^2-}$.

$\ce{Fe(OH)3}$ is insoluble, so it is written as a formula, not split into ions; $\ce{HNO3}$ is a strong acid written as $\ce{H+}$, and $\ce{NO3-}$ is a spectator. Choice A applies only to soluble strong bases; B splits the solid into ions and keeps a spectator $\ce{Fe^3+}$; C is a molecular equation.