Write the balanced molecular, complete ionic, and net ionic equations for each of the following acid-base reactions. (Type your answer using the format [NH4]+ for NH4+ and [Ni(CN)4]2- for Ni(CN)42-. Use the lowest possible coefficients.)

(a) HClO4(aq) + Mg(OH)2(s)
molecular equation, complete ionic, net ionic
(b) HCN(aq) + NaOH(aq)
molecular equation, complete ionic, net ionic
(c) HCl(aq) + NaOH(aq)
molecular,complete ionic, net ionic

These look easy enough to balance. What is your problem? What is it you don't understand? Follow the rule of

acid + base ==> salt + HOH

aqueous strontium hydroxide and hydroiodic acid net equation

(a)

Molecular equation: HClO4(aq) + Mg(OH)2(s) -> Mg(ClO4)2(aq) + H2O(l)

Complete ionic equation: 2H+(aq) + ClO4-(aq) + Mg2+(aq) + 2OH-(aq) -> Mg(ClO4)2(aq) + 2H2O(l)

Net ionic equation: 2H+(aq) + 2OH-(aq) -> 2H2O(l)

(b)

Molecular equation: HCN(aq) + NaOH(aq) -> NaCN(aq) + H2O(l)

Complete ionic equation: H+(aq) + CN-(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + CN-(aq) + H2O(l)

Net ionic equation: H+(aq) + OH-(aq) -> H2O(l)

(c)

Molecular equation: HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)

Complete ionic equation: H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + Cl-(aq) + H2O(l)

Net ionic equation: H+(aq) + OH-(aq) -> H2O(l)

(a) Let's start by writing the balanced molecular equation for the reaction between HClO4(aq) and Mg(OH)2(s):

HClO4(aq) + Mg(OH)2(s) → Mg(ClO4)2(aq) + H2O(l)

To write the complete ionic equation, we need to break down the aqueous compounds into their respective ions:

H+(aq) + ClO4-(aq) + Mg2+(aq) + 2OH-(aq) → Mg2+(aq) + 2ClO4-(aq) + 2H2O(l)

Next, we can cancel out the spectator ions that appear on both sides of the equation. In this case, the spectator ions are Mg2+ and ClO4-.

H+(aq) + OH-(aq) → H2O(l)

The resulting equation is the net ionic equation for the reaction.

(b) Now let's write the balanced molecular equation for the reaction between HCN(aq) and NaOH(aq):

HCN(aq) + NaOH(aq) → NaCN(aq) + H2O(l)

Breaking down the aqueous compounds into their respective ions, the complete ionic equation is:

H+(aq) + CN-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + CN-(aq) + H2O(l)

Again, we can cancel out the spectator ions, which are Na+ and CN-:

H+(aq) + OH-(aq) → H2O(l)

Therefore, the net ionic equation for this reaction is the same as the complete ionic equation.

(c) Lastly, let's write the balanced molecular equation for the reaction between HCl(aq) and NaOH(aq):

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Breaking down the aqueous compounds into their respective ions, the complete ionic equation is:

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + Cl-(aq) + H2O(l)

Once again, we can cancel out the spectator ions, which are Na+ and Cl-:

H+(aq) + OH-(aq) → H2O(l)

Thus, the net ionic equation for this reaction is the same as the complete ionic equation.