Copper(II) chloride and lead(II) nitrate react in aqueous solutions by double replacement. Write the balanced chemical equation, the overall ionic equation, and the net ionic equation for this reaction. If 11.95 g of copper(II) chloride react, what is the maximum amount of precipitate that could be formed?

First, let's identify the compounds and the products of the reaction. Copper(II) chloride is CuCl₂, and lead(II) nitrate is Pb(NO₃)₂, after the reaction the products will be PbCl₂ and Cu(NO₃)₂, the first one is an insoluble salt, which will precipitate, and the second one is a soluble salt. So, the balanced chemical equation will be:

CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)

The ionic equation is done by putting the ions that are presented when the compound ionization at the aqueous solution. The metals have their charged expressed in the name of the compound, and chloride and nitrate have charge -1:

Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)

The net ionic equation is the simplified ionic equation. So let's eliminate the ions that are presented on both sides of the equation:

2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)

The stoichiometry of the reaction is 1 mol of CuCl₂ for 1 mol of PbCl₂(the precipitate). The molar mass of the compounds are:

CuCl₂ = 134.452 g/mol

PbCl₂ = 278.106 g/mol

1 mol of CuCl₂ ------------ 1 mol of PbCl₂

Transforming in mass:

134.452 g of CuCl₂ ----------------- 278.106 g of PbCl₂

11.95 g of CuCl₂ ---------------- x

By a simple direct three rule:

134.452x = 3323.3667

x = 24.72 g of PbCl₂

throwdolbeau throwdolbeau · Answer 2 · 2022-03-10T11:56:36+01:00

Balanced chemical equation:

CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)

Overall ionic equation:

Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)

Net ionic equation:

2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)

The maximum amount of precipitate that could be formed is 24.72 g.

Balanced chemical reaction:

Copper(II) chloride is CuCl₂, and lead(II) nitrate is Pb(NO₃)₂, these react together with to form PbCl₂ and Cu(NO₃)₂. The balanced chemical equation will be:

CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)

The metals have their charged expressed in the name of the compound, and chloride and nitrate have charge -1. The ionic equation will be:

Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)

Net-ionic equation:

2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)

The stoichiometry of the reaction is 1 mol of CuCl₂ for 1 mol of PbCl₂(the precipitate). The molar mass of the compounds are:

CuCl₂ = 134.452 g/mol

PbCl₂ = 278.106 g/mol

1 mol of CuCl₂ ------------> 1 mol of PbCl₂

Converting to mass:

134.452 g of CuCl₂ ----------------- 278.106 g of PbCl₂

11.95 g of CuCl₂ ---------------- x g

134.452x = 3323.3667

x = 24.72 g of PbCl₂

The maximum amount of precipitate that could be formed is 24.72 g.

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