Q18E - Chemistry (OpenStax)

Question

The electrolysis of molten sodium chloride or of aqueous sodium chloride produces chlorine. Calculate the mass of chlorine produced from 3.00 kg sodium chloride in each case. You may wish to review the chapter on electrochemistry for relevant examples.

Step-by-Step Solution

Answer

The mass of \(m\left( {C{l_2}} \right) = 1.820kg\)

Step 1: The electrolysis of molten sodium chloride:

When sodium chloride is melted at high temperatures, it separates into sodium and chloride ions, allowing electrolysis to produce sodium atoms and chlorine gas.

Step 2: Calculate the number of moles of sodium chloride:

\(n({\rm{NaCl}}) = \frac{{m({\rm{NaCl}})}}{{{\rm{M}}({\rm{NaCl}})}}{\rm{  }}\)

\(n({\rm{NaCl}}) = \frac{{3 \cdot {{10}^3}\;{\rm{g}}}}{{58.44\;{\rm{g}}/{\rm{mol}}}}{\rm{ }}\)

\(n({\rm{NaCl}}) = 51.33\;{\rm{mol }}\)

\({\rm{The reaction : }}\)

\(2{\rm{NaCl}}({\rm{aq}}) + 2{{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to \)\(2{\rm{NaOH}}({\rm{Oq}}) + {\rm{C}}{{\rm{l}}_2}(\;{\rm{g}}) + {{\rm{H}}_2}(\;{\rm{g}})\)

Step 3: To calculate the number of moles:

To calculate the number of moles produced of \({\rm{Cl}}{l_2},\)multiply the number of moles of \({\rm{NaCl with the}}\)ratio of their stoichiometric coefficients (stoichiometric ratio).

\(\frac{{\nu ({\rm{NaCl}})}}{{\nu \left( {{\rm{Cl}}{l_2}} \right)}} = \frac{2}{1}{\rm{ }}\)

\(n\left( {{\rm{Cl}}{l_2}} \right) = n({\rm{NaCl}}) \cdot \frac{{\nu ({\rm{Cl}}2)}}{{\nu ({\rm{NaCl}})}}{\rm{  }}\)

\(n\left( {{\rm{Cl}}{l_2}} \right) = 51.33\;{\rm{mol}} \cdot \frac{1}{2}{\rm{ }}\)

\(n\left( {{\rm{Cl}}{l_2}} \right) = 25.67\;{\rm{mol}}\)

Step 4: Calculate the mass of \(\left( {C{l_2}} \right)\) :

Finally, calculate the mass of\(\left( {C{l_2}} \right)\):

\(m\left( {C{l_2}} \right) = n\left( {C{l_2}} \right) \cdot M\left( {C{l_2}} \right)\)

\(m\left( {C{l_2}} \right) = 25.67\;{\rm{mol}} \cdot 70.90\;{\rm{g}}/{\rm{mol}}\)

\(m\left( {C{l_2}} \right) = 1.820\;{\rm{kg}}\)