Question

Write a balanced equation for Sodium thiosulphate is reacted with dilute Hydrochloric acid.

Answer 1

Step 1: Reacting to the given molecules:

• When Hydrochloric acid reacts with Sodium thiosulphate to form Sodium Chloride, Sulphur Dioxide, Sulphur and water.
• The unbalanced chemical equation can be written as

1. $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{NaCl}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{SO}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{}\mathbf{S}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sulphur}\mathbf{} \\ \mathbf{Thiosulphate}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Dioxide} \end{gathered}$$

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: $\mathbf{2}$ Sodium atoms from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$, $\mathbf{2}$ Sulphur atoms from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$, $\mathbf{3}$ atoms from Oxygen from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$, $\mathbf{1}$ atom of Hydrogen from $\mathbf{HCl}$and $\mathbf{1}$ Chlorine atom from $\mathbf{HCl}$
• On the product side: $\mathbf{1}$ Sodium atom from $\mathbf{NaCl}$, $\mathbf{1}$ Chlorine atom from $\mathbf{NaCl}$, $\mathbf{2}$ Sulphur atom from ${\mathbf{SO}}_{\mathbf{2}}$ and $\mathbf{S}$, $\mathbf{2}$ Hydrogen atoms from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ and $\mathbf{3}$ Oxygen atoms from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ and ${\mathbf{SO}}_{\mathbf{2}}$

Step 3: Balancing the reactant molecules by multiplying $\mathbf{2}$ by $\mathbf{HCl}$ and $\mathbf{2}$ by $\mathbf{NaCl}$

• The chemical reaction is as follows:
• $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{2}\mathbf{}\mathbf{NaCl}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{SO}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{S}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sulphur}\mathbf{} \\ \mathbf{Thiosulphate}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Dioxide} \end{gathered}$$
• The total number of atoms of each element on the reactant side and the product side must be compared.
• On the reactant side: $\mathbf{2}$ Sodium atoms from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$, $\mathbf{2}$ Sulphur atoms from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$, $\mathbf{3}$ atoms from Oxygen from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$, $\mathbf{2}$ atom of Hydrogen from $\mathbf{HCl}$and $\mathbf{2}$ Chlorine atom from $\mathbf{HCl}$
• On the product side: $\mathbf{2}$ Sodium atom from $\mathbf{NaCl}$, $\mathbf{2}$ Chlorine atom from $\mathbf{NaCl}$, $\mathbf{2}$ Sulphur atom from ${\mathbf{SO}}_{\mathbf{2}}$ and $\mathbf{S}$, $\mathbf{2}$ Hydrogen atoms from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ and $\mathbf{3}$ Oxygen atoms from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ and ${\mathbf{SO}}_{\mathbf{2}}$.
• After multiplication, we have an equal number of molecules on both sides of the reaction.

Step 4: Balanced chemical equation is as follows:

• The chemical reaction is as follows:

$$\begin{gathered} \mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{2}\mathbf{}\mathbf{NaCl}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{SO}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{S}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sulphur}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sulphur}\mathbf{} \\ \mathbf{Thiosulphate}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Dioxide} \end{gathered}$$

Hence, the equation is balanced.