Question

Chlorine gas was first prepared in 1774 by \(\mathrm{C}\). W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is $$ \begin{aligned} \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) & \longrightarrow \\ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) &+\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g) \end{aligned} $$ Balance this equation.

Short answer

The balanced equation is: \[ 2\mathrm{NaCl}(a q) +\mathrm{H}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]

Step-by-step solution

1. Write down the given unbalanced equation

The given unbalanced equation is: \[ \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]

2. Balance chlorine atoms

Start by balancing Cl atoms since they are present in more than one molecule. We have 1 Cl atom on the left side and 3 Cl atoms on the right side (1 from Na2SO4 and 2 from MnCl2). To balance Cl atoms, add a coefficient of 2 in front of NaCl: \[ 2\mathrm{NaCl}(a q) +\mathrm{H}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]

3. Balance sodium atoms

Next, balance Na atoms. We now have 2 Na atoms on both the left and right sides. So, no change is needed for Na atoms.

4. Balance manganese atoms

Now, balance Mn atoms. There is 1 Mn atom on both the left and right sides. So, no change is needed for Mn atoms.

5. Balance oxygen atoms

Balance O atoms. We have 6 O atoms on the left side (4 from H2SO4 and 2 from MnO2) and 5 O atoms on the right side (4 from Na2SO4 and 1 from H2O). To balance O atoms, add a coefficient of 2 in front of H2O: \[ 2\mathrm{NaCl}(a q) +\mathrm{H}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]

6. Balance hydrogen atoms

Lastly, balance H atoms. We now have 4 H atoms on the left side (2 from H2SO4) and 4 H atoms on the right side (4 from 2 H2O). So, no change is needed for H atoms.

7. Check if the equation is balanced

Verify if all elements are balanced in the equation: - Na: 2 atoms on both sides, - Cl: 2 atoms on both sides, - H: 4 atoms on both sides, - S: 1 atom on both sides, - Mn: 1 atom on both sides, - O: 6 atoms on both sides. Since all elements are balanced, the balanced equation is: \[ 2\mathrm{NaCl}(a q) +\mathrm{H}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+2\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g) \]

Key concepts

Chlorine Gas Production

Chlorine gas was initially produced in 1774 by a scientist named Carl Wilhelm Scheele. This process involved a chemical reaction where sodium chloride (NaCl), commonly known as table salt, was oxidized using manganese(IV) oxide (MnO₂). Through this reaction, chlorine gas (Cl₂) was released as a product. Chlorine gas production is significant due to its vast applications in industry, particularly for disinfecting water and producing other chemicals. For Scheele's experiment, the essential chemicals—sodium chloride, sulfuric acid (H₂SO₄), and manganese(IV) oxide—combined to yield products like manganese chloride (MnCl₂), sodium sulfate (Na₂SO₄), water, and the highly desired chlorine gas itself. Understanding this chemical process is vital for appreciating how various industrial compounds are synthesized. Recognizing each component's role in producing chlorine gas helps us appreciate the complexity and precision involved in chemical manufacturing.

Oxidation Reaction

An oxidation reaction is crucial in chemistry as it involves the transfer of electrons between substances. In the context of chlorine gas production, the exchange of electrons is central to the process. Manganese(IV) oxide acts as an oxidizing agent, meaning it accepts electrons from sodium chloride during the reaction. As a result, the chemical bonds within sodium chloride are broken, leading to the formation of chlorine gas.

In simple terms, oxidation often involves gaining oxygen or losing hydrogen for a particular substance. Conversely, reduction involves losing oxygen or gaining hydrogen. This reaction showcases a classic redox (reduction-oxidation) process, where manganese is reduced, and chloride ions from the sodium chloride are oxidized. By understanding how oxidation reactions function, students can better grasp how compounds change their structure and composition during chemical reactions.

Balancing Atoms

Balancing chemical equations is essential to reflect the principle of the conservation of mass. In any chemical reaction, the total number of each type of atom must be the same on both the reactant and product sides.

Consider the equation for chlorine gas production: unbalanced (\[\mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q) +\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2}\mathrm{O}(l)+\mathrm{Cl}_{2}(g)\]).

To balance it, one must ensure that each side has an equivalent number of sodium (Na), chloride (Cl), sulfur (S), manganese (Mn), oxygen (O), and hydrogen (H) atoms. This is achieved by adjusting the coefficients before the compounds:

• For Cl, a coefficient of 2 is necessary for NaCl to mirror the Cl atoms in MnCl₂.
• Balancing other components like manganese (Mn) or sulfur (S) might not require changes, as they are naturally balanced in this particular reaction.
• Accounting for oxygen and hydrogen could entail changing coefficients to ensure both sides boast equal numbers.

Such a balanced equation, by adhering to these rules, confirms the integrity of mass conservation in chemical transformations.