Question

Predict whether a reaction occurs, and write balanced complete and net ionic equations:(a) Iron (III) chloride (aq) + cesium phosphate (aq) →(b) Sodium hydroxide (aq) + cadmium nitrate (aq) →(c) Magnesium bromide (aq) + potassium acetate (aq) →(d) Silver sulfate (aq) + barium chloride (aq) →(e) Sodium sulfate (aq) + strontium nitrate (aq) →

Answer 1

Answer : The balanced complete and net ionic equations are written below.

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a) The balanced molecular equation will be,

$FeCl_3(aq)+Cs_3PO_4(aq)\rightarrow 3CsCl(aq)+FePO_4(s)$

The complete ionic equation in separated aqueous solution will be,

$Fe^{3+}(aq)+3Cl^{-}(aq)+3Cs^{+}(aq)+PO_4^{3-}(aq)\rightarrow 3Cs^{+}(aq)+3Cl^{-}(aq)+FePO_4(s)$

In this equation the species present are, $3Cl^{-}\text{ and }Cs^{+}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Fe^{3+}(aq)+PO_4^{3-}(aq)\rightarrow FePO_4(s)$

(b) The balanced molecular equation will be,

$2NaOH(aq)+Cd(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Cd(OH)_2(s)$

The complete ionic equation in separated aqueous solution will be,

$2Na^{+}(aq)+2OH^{-}(aq)+Cd^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^{+}(aq)+2NO_3^{-}(aq)+Cd(OH)_2(s)$

In this equation the species present are, $Na^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Cd^{2+}(aq)+2OH^{-}(aq)\rightarrow Cd(OH)_2(s)$

(c) The balanced molecular equation will be,

$MgBr_2(aq)+2CH_3COOK(aq)\rightarrow (CH_3COO)_2Mg(aq)+2KBr(aq)$

The complete ionic equation in separated aqueous solution will be,

$Mg^{2+}(aq)+2Br^{-}(aq)+2K^{+}(aq)+2CH_3COO^{-}(aq)\rightarrow Mg^{2+}(aq)+2CH_3COO^{-}(aq)+2Br^{-}(aq)+2K^{+}(aq)$

In this equation all the species are the spectator ions.

Thus, there is no net ionic equation.

(d) The balanced molecular equation will be,

$Ag_2SO_4(aq)+BaCl_2(aq)\rightarrow 2AgCl(s)+BaSO_4(s)$

The complete ionic equation in separated aqueous solution will be,

$2Ag^{+}(aq)+SO_4^{2-}(aq)+Ba^{2+}(aq)+2Cl^{-}(aq)\rightarrow 2AgCl(s)+BaSO_4(s)$

In this equation no species are the spectator ions.

Thus, there is no net ionic equation.

(e) The balanced molecular equation will be,

$Na_2SO_4(aq)+Sr(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+SrSO_4(s)$

The complete ionic equation in separated aqueous solution will be,

$2Na^{+}(aq)+SO_4^{2-}(aq)+Sr^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^{+}(aq)+2NO_3^{-}(aq)+SrSO_4(s)$

In this equation the species present are, $Na^{+}\text{ and }NO_3^{-}$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Sr^{2+}(aq)+SO_4^{2-}(aq)\rightarrow SrSO_4(s)$