Question

Chlorine gas was first prepared in 1774 by C. W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is $$\mathrm{NaCl}(aq)+{\mathrm{H}}_2{\mathrm{SO}}_4(aq)+{\mathrm{MnO}}_2(s)⟶{\mathrm{Na}}_2{\mathrm{SO}}_4(aq)+{\mathrm{MnCl}}_2(aq)+{\mathrm{H}}_2\mathrm{O}(l)+{\mathrm{Cl}}_2(g)$$ Balance this equation.

Short answer

The balanced equation is: $2\mathrm{NaCl}(aq)+{\mathrm{H}}_2{\mathrm{SO}}_4(aq)+{\mathrm{MnO}}_2(s)⟶{\mathrm{Na}}_2{\mathrm{SO}}_4(aq)+{\mathrm{MnCl}}_2(aq)+2{\mathrm{H}}_2\mathrm{O}(l)+{\mathrm{Cl}}_2(g)$

Step-by-step solution

Write the unbalanced equation

The unbalanced equation is given as: NaCl(aq) + H2SO4(aq) + MnO2(s) -> Na2SO4(aq) + MnCl2(aq) + H2O(l) + Cl2(g)

Count the number of atoms for each element on both sides of the equation

On the left side of the equation (reactants), we have: - Sodium (Na): 1 - Chlorine (Cl): 1 - Hydrogen (H): 2 - Sulfur (S): 1 - Manganese (Mn): 1 - Oxygen (O): 6 On the right side of the equation (products), we have: - Sodium (Na): 2 - Chlorine (Cl): 2 - Hydrogen (H): 2 - Sulfur (S): 1 - Manganese (Mn): 1 - Oxygen (O): 5

Balance the equation

First, we'll balance the sodium atoms by placing a 2 in front of NaCl: 2NaCl(aq) + H2SO4(aq) + MnO2(s) -> Na2SO4(aq) + MnCl2(aq) + H2O(l) + Cl2(g) Next, we can see that the chlorine atoms are also balanced, but the number of oxygen atoms on the product side is still less than on the reactant side. To fix this issue, we'll add a coefficient of 2 in front of H2O: 2NaCl(aq) + H2SO4(aq) + MnO2(s) -> Na2SO4(aq) + MnCl2(aq) + 2H2O(l) + Cl2(g) Now, counting the atoms again, we have: On the left side of the equation (reactants): - Sodium (Na): 2 - Chlorine (Cl): 2 - Hydrogen (H): 2 - Sulfur (S): 1 - Manganese (Mn): 1 - Oxygen (O): 6 On the right side of the equation (products): - Sodium (Na): 2 - Chlorine (Cl): 2 - Hydrogen (H): 2 - Sulfur (S): 1 - Manganese (Mn): 1 - Oxygen (O): 6 Both sides match, so our balanced chemical equation is: 2NaCl(aq) + H2SO4(aq) + MnO2(s) -> Na2SO4(aq) + MnCl2(aq) + 2H2O(l) + Cl2(g)

Key concepts

Oxidation Reactions

Oxidation reactions involve the transfer of electrons between substances, often resulting in the alteration of oxidation states. When balancing chemical equations involving oxidation, it's important to ensure that the number of electrons lost is equal to the number of electrons gained. This maintains the charge neutrality of the reaction.
In the preparation of chlorine gas, the chemical reaction involves sodium chloride (NaCl), which contains chlorine that gets oxidized.

• Here, chlorine begins as chloride ions (Cl^-) in the reactants.
• It is oxidized to form chlorine gas (Cl₂), where each chlorine atom achieves a neutral charge.

Oxidation reactions are crucial for many industrial processes, including the synthesis of chemicals, waste treatment, and energy conversions in batteries.
Understanding oxidation reactions allows chemists to predict the behavior of reactants and manipulate them to achieve desired products. This insight is especially vital in processes like chlorine gas preparation.

Chlorine Gas Preparation

Chlorine gas is a vital industrial chemical used in a variety of applications such as disinfection, textile processing, and the production of plastics like PVC. The preparation of chlorine gas through the reaction of sodium chloride, sulfuric acid, and manganese(IV) oxide, as demonstrated by Scheele, highlights an early method of gas preparation.
To understand this process, consider the role of each reactant:

• Sodium chloride ( NaCl): provides the chlorine source.
• Sulfuric acid ( H₂SO₄): acts as a catalyst and source of protons (H⁺).
• Manganese(IV) oxide ( MnO₂): serves as an oxidizing agent to facilitate the release of Cl₂ gas from Cl^- ions.

By carefully balancing such chemical reactions, chemists are able to optimize the production of chlorine gas, minimizing waste and maximizing efficiency. A balanced equation ensures all atoms are accounted for, following the law of conservation of mass.

Stoichiometry

Stoichiometry is the practice of relating the quantities of reactants and products in a chemical reaction. This field of chemistry is foundational for predicting the amounts of substances consumed and created in reactions. The balanced chemical equation obtained in stoichiometry provides a mole ratio, which is pivotal for these calculations.
In the balanced reaction for chlorine gas preparation, the coefficients in the balanced equation:$$2\text{NaCl}(aq)+{\text{H}}_2{\text{SO}}_4(aq)+{\text{MnO}}_2(s)\to {\text{Na}}_2{\text{SO}}_4(aq)+{\text{MnCl}}_2(aq)+2{\text{H}}_2\text{O}(l)+{\text{Cl}}_2(g)$$

• "2" in front of NaCl shows that 2 moles of NaCl react relative to other substances.
• This mole ratio helps calculate how much MnO₂ is needed or how much Cl₂ can be produced from given amounts of reactants.
• It also informs how reagents are scaled up or down for larger or smaller reactions, ensuring efficient production practices.

Mastery of stoichiometry enables scientists and engineers to predict outcomes accurately, ensuring safety and cost efficiency in both laboratory and industrial settings. Proper stoichiometric calculations are essential for the success of chemical manufacturing, fuel processing, and various laboratory analyses.