Question

Liquid carbon disulfide reacts with oxygen gas, producing carbon dioxide gas and sulfur dioxide gas.

Short answer

The balanced chemical equation for the reaction between liquid carbon disulfide and oxygen gas is: \( 2CS_2 (l) + 5O_2 (g) \rightarrow 2CO_2 (g) + 4SO_2 (g) \).

Step-by-step solution

Write the unbalanced chemical equation

First, let's write the unbalanced chemical equation for the given reaction: \( CS_2 (l) + O_2 (g) \rightarrow CO_2 (g) + SO_2 (g) \) In this equation, the abbreviations (l), (g), and (s) represent the states of the compounds: liquid, gas, and solid, respectively.

Count the atoms

Now, let's count the number of atoms on both sides of the equation: - Left side (Reactants): - 1 Carbon atom - 2 Sulfur atoms - 2 Oxygen atoms - Right side (Products): - 1 Carbon atom - 1 Sulfur atom - 2 Oxygen atoms in CO2 molecule - 2 Oxygen atoms in SO2 molecule As we can see, the number of sulfur and oxygen atoms is not balanced between the reactants and products.

Balance the chemical equation

In order to balance the chemical equation, we may need to adjust the coefficients (the numbers in front of the chemical formulas) until the number of atoms for each element is equal on both sides. 1. Balance sulfur atoms by placing a coefficient of 2 in front of SO2: \( CS_2 (l) + O_2 (g) \rightarrow CO_2 (g) + 2SO_2 (g) \) - Now we have 2 sulfur atoms on both sides. 2. Balance the oxygen atoms by adjusting the coefficient of O2: \( CS_2 (l) + 5/2O_2 (g) \rightarrow CO_2 (g) + 2SO_2 (g) \) However, having a fraction as a coefficient is not preferred in chemical equations. To fix this issue, we need to multiply the whole equation by 2 to obtain whole number coefficients. 3. Multiply the entire equation by 2 to eliminate the fraction: \( 2CS_2 (l) + 5O_2 (g) \rightarrow 2CO_2 (g) + 4SO_2 (g) \) Now, the chemical equation is balanced with the equal number of atoms of each element on both sides.

Verify the balanced equation

Finally, verify that the balanced equation is correct by counting the atoms on both sides: - Left side (Reactants): - 2 Carbon atoms - 4 Sulfur atoms - 10 Oxygen atoms - Right side (Products): - 2 Carbon atoms - 4 Sulfur atoms - 10 Oxygen atoms As we can see, the number of atoms for each element is now equal on both sides, which means our balanced chemical equation is correct: \( 2CS_2 (l) + 5O_2 (g) \rightarrow 2CO_2 (g) + 4SO_2 (g) \)

Key concepts

Understanding Chemical Equations

A chemical equation is similar to a recipe for a chemical reaction. It provides the necessary ingredients and their proportions to create a certain reaction. When writing a chemical equation, substances that react (called reactants) are written on the left side, while the new substances formed (referred to as products) are written on the right side. The reactants and products are represented by their chemical formulas, and states of matter are denoted by abbreviations such as (l) for liquid, (g) for gas, and (s) for solid, as seen in the exercise.

The process begins with an unbalanced chemical equation, which may not accurately represent the reaction due to different quantities of atoms on each side. As indicated in the provided step-by-step solution, balancing starts by counting and comparing the number of atoms for each element on both sides of the equation. The ultimate goal is to demonstrate that matter is neither created nor destroyed during a chemical reaction, which is known as the Law of Conservation of Mass.

Incorrectly balanced equations could lead to a flawed understanding of how much of each reactant is needed and what amount of product will be formed. Students are encouraged to always double-check their balanced equations to ensure accuracy.

The Crucial Role of Stoichiometry

Stoichiometry is a key concept in chemistry, which involves the calculation of relative quantities of reactants and products in chemical reactions. It is founded on the principle that the quantities of substances are connected by ratios determined by the balanced equation.

In the exercise, the stoichiometric coefficients (numbers placed in front of the chemical formulas) are adjusted to obey the Law of Conservation of Mass, reflecting the correct amounts of reactants and products. The solution demonstrates this adjustment by changing the coefficients to ensure that there are equal numbers of each type of atom on both sides of the equation.

One of the most common challenges for students is dealing with coefficients that are fractions. As seen with the reaction involving carbon disulfide and oxygen, coefficients should ultimately be whole numbers. If students encounter fractions, they should follow the step of multiplying the entire equation by the appropriate factor to achieve this, thereby mastering the nuances of stoichiometry.

The Law of Conservation of Mass

According to the Law of Conservation of Mass, mass is neither created nor destroyed in a chemical reaction. Thus, a chemical equation must reflect this principle by showing that the same number of each type of atom exists on both sides of the reaction equation. This law is fundamental to the science of chemistry and serves as the basis for balancing chemical equations.

In practice, to balance an equation, such as the provided example of carbon disulfide and oxygen reacting, one might have to manipulate the coefficients to ensure that each atom's mass is conserved. In the final verification step, as seen in the solution, you check to ensure that the number of atoms of each element is identical on the reactant and product sides of the balanced equation.

This verification upholds the Law of Conservation of Mass and confirms the accuracy of the stoichiometric balance achieved in the chemical equation. Encouraging students to internalize this law helps to deepen their understanding of chemical reactions and the predictable nature of matter during these interactions.