Question

• Write a balanced equation red Lead is warmed with concentrated Hydrochloric acid.

Answer 1

Step 1: Reacting to the given molecules:

• The chemical formula of Hydrochloric acid is $\mathbf{HCl}$. It reacts with Red Lead to form Lead Chloride, Chlorine and Water.
• The unbalanced chemical equation can be written as:
• $$\begin{gathered} \mathbf{}{\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{PbCl}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{Red}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Lead}\mathbf{} \\ \mathbf{}\mathbf{Lead}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride} \end{gathered}$$

Step 2: Balancing the chemical equation:

• The total number of atoms of each element on the reactant side and the product side must be compared.
• On the reactant side: $\mathbf{3}$ Lead atoms from ${\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$, $\mathbf{4}$ Oxygen atoms from ${\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$, $\mathbf{1}$ atom of Hydrogen from HCl and $\mathbf{1}$ Chlorine atom from $\mathbf{HCl}$
• On the product side: $\mathbf{1}$ Lead atom from ${\mathbf{PbCl}}_{}$, $\mathbf{2}$ Hydrogen atoms from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$, $\mathbf{3}$ Chlorine atoms from ${\mathbf{PbCl}}_{}$and ${\mathbf{Cl}}_{\mathbf{2}}$ and $\mathbf{1}$ Oxygen atom from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

Step3: Balancing the reactant molecules by multiplying $\mathbf{2}$ by ${\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$, and $\mathbf{6}$ by ${\mathbf{PbCl}}_{}$,

The chemical reaction is as follows:

• $$\begin{gathered} \mathbf{2}{\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{6}\mathbf{PbCl}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{Cl}}_{\mathbf{2}\mathbf{}}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{Red}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Lead}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{Lead}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride} \end{gathered}$$
• On the reactant side: $\mathbf{6}$ Lead atoms from ${\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$, $\mathbf{8}$ Oxygen atoms from ${\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$, $\mathbf{1}$ atom of Hydrogen from HCl and $\mathbf{1}$ Chlorine atom from $\mathbf{HCl}$.
• On the product side: $\mathbf{6}$ Lead atom from ${\mathbf{PbCl}}_{}$, $\mathbf{2}$ Hydrogen atoms from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$, $\mathbf{8}$ Chlorine atoms from ${\mathbf{PbCl}}_{}$ and ${\mathbf{Cl}}_{\mathbf{2}}$ and $\mathbf{1}$ Oxygen atom from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$.

Step 4: Balancing the reactant molecules by multiplying $\mathbf{16}$ by $\mathbf{HCl}$, $\mathbf{8}$ by ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ and $\mathbf{2}$ by ${\mathbf{Cl}}_{\mathbf{2}}$

• The chemical reaction is as follows:
• $$\begin{gathered} \mathbf{2}{\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{16}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{6}\mathbf{PbCl}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{8}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{5}{\mathbf{Cl}}_{\mathbf{2}\mathbf{}}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{Red}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Lead}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{Lead}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride} \end{gathered}$$
• On the reactant side: $\mathbf{6}$ Lead atoms from ${\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$, $\mathbf{8}$ Oxygen atoms from ${\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}$, $\mathbf{16}$ an atom of Hydrogen from HCl and $\mathbf{16}$ Chlorine atom from $\mathbf{HCl}$
• On the product side: $\mathbf{6}$ Lead atom from ${\mathbf{PbCl}}_{}$, $\mathbf{16}$ Hydrogen atoms from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$, $\mathbf{16}$ Chlorine atoms from ${\mathbf{PbCl}}_{}$ and ${\mathbf{Cl}}_{\mathbf{2}}$ and $\mathbf{8}$ Oxygen atom from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$.
• After multiplication, we have an equal number of molecules on both sides of the reaction

Step 5: Balanced chemical equation is as follows:

1. Therefore, the balanced equation is
2. $$\begin{gathered} \mathbf{2}{\mathbf{Pb}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{16}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{6}\mathbf{PbCl}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{8}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{5}{\mathbf{Cl}}_{\mathbf{2}\mathbf{}}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{Red}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Lead}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{Lead}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride} \end{gathered}$$

Hence, the equation is balanced.