Chapter 8, Problem 8.65QA

Interpretation Introduction

To find

We need to complete and balance the given chemical equations and write the net ionic equations.

Answer to Problem 8.65QA

Solution

a) Balanced equation: $Pb{(NO}_3{)}_{2\left(aq\right)}+{Na}_2{SO}_{4\left(aq\right)}\to {PbSO}_{4\left(s\right)}+2{NaNO}_{3\left(aq\right)}$

Net ionic equation: ${Pb}_{\left(aq\right)}^{2+}+{SO}_{4\left(aq\right)}^{2-}\to {PbSO}_{4\left(s\right)}$

b) No precipitation reaction occurs.

c) Balanced equation: ${FeCl}_{2\left(aq\right)}+{Na}_2{S}_{\left(aq\right)}\to {FeS}_{\left(s\right)}+{2NaCl}_{\left(aq\right)}$

Net ionic equation: ${Fe}_{\left(aq\right)}^{2+}+S_{\left(aq\right)}^{2-}\to {FeS}_{\left(s\right)}$

d) Balanced equation: ${MgSO}_{4\left(aq\right)}+{BaCl}_{2\left(aq\right)}\to {MgCl}_{2\left(aq\right)}+{BaSO}_{4\left(s\right)}$

Net ionic equation: ${Ba}_{\left(aq\right)}^{2+}+{SO}_{4\left(aq\right)}^{2-}\to {BaSO}_{4\left(s\right)}$

Explanation of Solution

To balance the chemical reactions, we use the law of conservation of matter. According to the law of conversion of matter, the number of atoms present on the reactants sides is equal to the number of atoms present on the product side.

To write a net ionic equation for a reaction in which the solid precipitates, we begin with a molecular equation, and then write the corresponding total ionic equation that includes all of the individual ions present in the solutions of soluble reactants and products. Eliminating the spectator ions yield the net ionic equation.

a) The complete chemical reaction is

$Pb{(NO}_3{)}_{2\left(aq\right)}+{Na}_2{SO}_{4\left(aq\right)}\to {PbSO}_{4\left(s\right)}+{NaNO}_{3\left(aq\right)}$

Atoms: $1Pb+2N+10O+1S+2Na\to 1Pb+7O+1S+1Na+1N$

The lead and sulfur atoms are already balanced. There are 2 Na on the left but only one on the right. To balance sodium, we place a coefficient of 2 in front of ${NaNO}_3$

$Pb{(NO}_3{)}_{2\left(aq\right)}+{Na}_2{SO}_{4\left(aq\right)}\to {PbSO}_{4\left(s\right)}+2{NaNO}_{3\left(aq\right)}$

Atoms: $1Pb+2N+10O+1S+2Na\to 1Pb+10O+1S+2Na+2N$

Balanced equation: $Pb{(NO}_3{)}_{2\left(aq\right)}+{Na}_2{SO}_{4\left(aq\right)}\to {PbSO}_{4\left(s\right)}+2{NaNO}_{3\left(aq\right)}$

According the solubility rules, all nitrate and sulfates are soluble in water except sulfate of  Pb2+ cations; they are insoluble. So, they are written as separate ions in a total ionic equation. However, solid lead sulfate is written as ${PbSO}_{4\left(s\right)}$.

${Pb}_{\left(aq\right)}^{2+}+2{NO}_{3\left(aq\right)}^{-}+{2Na}_{\left(aq\right)}^{+}+{SO}_{4\left(aq\right)}^{2-}\to {PbSO}_{4\left(s\right)}+{2Na}_{\left(aq\right)}^{+}+2{NO}_{3\left(aq\right)}^{-}$

Eliminating the ${Na}_{\left(aq\right)}^{+}$ and ${NO}_{3\left(aq\right)}^{-}$ spectator ions yield the net ionic equation

${Pb}_{\left(aq\right)}^{2+}+{SO}_{4\left(aq\right)}^{2-}\to {PbSO}_{4\left(s\right)}$

b) When ${NiCl}_{2 }and {NH}_4{NO}_3$ combines they form Nickel nitrate and because all nitrate compounds are soluble, no precipitation reaction occurs between ${NiCl}_{2 }and {NH}_4{NO}_3$ .

${NiCl}_{2\left(aq\right)}+{{NH}_{4}NO}_{3\left(aq\right)}\to Ni{(NO}_3{)}_{2\left(aq\right)}+{NH}_4{Cl}_{\left(aq\right)}$

Atoms: $1Ni+2Cl+2 N+4H+3O\to 1 Ni+1 Cl+3 N+4 H+6 O$

Balance the Cl, H and O atoms to both side. Thus the complete balanced reaction is

${NiCl}_{2\left(aq\right)}+2 {{NH}_{4}NO}_{3\left(aq\right)}\to Ni{(NO}_3{)}_{2\left(aq\right)}+2{NH}_4{Cl}_{\left(aq\right)}$

Atoms: $1 Ni+2 Cl+4 N+8 H+6 O\to 1 Ni+2 Cl+4 N+8 H+6 O$

According to solubility rules, all nitrates and NH₄⁺ ions are soluble in water, so this will not form a precipitation reaction.

c) The complete chemical reaction is

${FeCl}_{2\left(aq\right)}+{Na}_2{S}_{\left(aq\right)}\to {FeS}_{\left(s\right)}+{NaCl}_{\left(aq\right)}$

Atoms: $1Fe+2Cl+2Na+1S\to 1Fe+1S+1Na+1Cl$

Iron and sulfur atoms are already balanced, so let’s focus on sodium and chlorine. There are 2 Na on the left but only one on the right. To balance sodium, we place a coefficient of 2 in front of $NaCl$ and recalculate the distribution of atoms on both sides.

${FeCl}_{2\left(aq\right)}+{Na}_2{S}_{\left(aq\right)}\to {FeS}_{\left(s\right)}+2{NaCl}_{\left(aq\right)}$

Atoms: $1Fe+2Cl+2Na+1S\to 1Fe+1S+2Na+2Cl$

Balanced equation: ${FeCl}_{2\left(aq\right)}+{Na}_2{S}_{\left(aq\right)}\to {FeS}_{\left(s\right)}+2{NaCl}_{\left(aq\right)}$

The reactants are soluble ionic compounds, so they are written as separate ions in a total ionic equation, as is soluble $NaCl$. However, solid ferrous sulfate is written as ${FeS}_{\left(s\right)}$

${Fe}_{\left(aq\right)}^{2+}+{2Cl}_{\left(aq\right)}^{-}+{2Na}_{\left(aq\right)}^{+}{S}_{\left(aq\right)}^{2-}\to {FeS}_{\left(s\right)}+{2Cl}_{\left(aq\right)}^{-}+{2Na}_{\left(aq\right)}^{+}$

Eliminating the spectator ions ${Cl}_{\left(aq\right)}^{-}$ and ${Na}_{\left(aq\right)}^{+}$ yield the net ionic equation,

Net ionic equation: ${Fe}_{\left(aq\right)}^{2+}+S_{\left(aq\right)}^{2-}\to {FeS}_{\left(s\right)}$

d) The complete chemical reaction is

${MgSO}_{4\left(aq\right)}+{BaCl}_{2\left(aq\right)}\to {MgCl}_{2\left(aq\right)}+{BaSO}_{4\left(s\right)}$

Atom: $1Mg+1S+1Ba+2Cl+4O\to :\mathrm{ }1Mg+1S+1Ba+2Cl+4O$

All the atoms are already balanced. So the entire balanced equation is

${MgSO}_{4\left(aq\right)}+{BaCl}_{2\left(aq\right)}\to {MgCl}_{2\left(aq\right)}+{BaSO}_{4\left(s\right)}$

The reactants are soluble ionic compounds, so they are written as separate ions in a total ionic equation, as is soluble ${MgCl}_2$. However, solid Barium sulfate is written as ${BaSO}_4$

${Ba}_{\left(aq\right)}^{2+}{+ Mg}_{\left(aq\right)}^{2+}+{2Cl}_{\left(aq\right)}^{-}+{SO}_{4\left(aq\right)}^{2-}\to {BaSO}_{4\left(s\right)}{+ Mg}_{\left(aq\right)}^{2+}+{2Cl}_{\left(aq\right)}^{-}$

Eliminating the spectator ions ${Mg}_{\left(aq\right)}^{2+}$ and ${Cl}_{\left(aq\right)}^{-}$ yield the net ionic equation,

${Ba}_{\left(aq\right)}^{2+}+{SO}_{4\left(aq\right)}^{2-}\to {BaSO}_{4\left(s\right)}$

Conclusion

The reactions are balanced according to  the law of conservation of matter and net ionic reactions determined from the solubility rules.