Write the balanced equation for the combustion of dimethylsulfide, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S},\) in an abundant supply of oxygen.

Short Answer

Expert verified
The balanced equation for the combustion of dimethylsulfide is \[2\left(\mathrm{CH}_{3}\right)_{2}S + 9O_{2} \rightarrow 4CO_{2} + 6H_{2}O + 2SO_{2}.\]

Step by step solution

01

Write the Unbalanced Equation for Combustion

Start by writing the reactants and products of the combustion of dimethylsulfide. The reactants are dimethylsulfide \(\left(\mathrm{CH}_{3}\right)_{2}S\) and oxygen (O2), and the products are carbon dioxide (CO2), water (H2O), and sulfur dioxide (SO2). The unbalanced equation is: \[\left(\mathrm{CH}_{3}\right)_{2}S + O_{2} \rightarrow CO_{2} + H_{2}O + SO_{2}.\]
02

Balance Carbon Atoms

Balance the carbon atoms first. Each molecule of dimethylsulfide has two carbon atoms, so you will need two molecules of CO2 to balance the carbon atoms: \[\left(\mathrm{CH}_{3}\right)_{2}S + O_{2} \rightarrow 2CO_{2} + H_{2}O + SO_{2}.\]
03

Balance Hydrogen Atoms

Next, balance the hydrogen atoms. There are six hydrogen atoms in one molecule of dimethylsulfide, thus three molecules of water are needed: \[\left(\mathrm{CH}_{3}\right)_{2}S + O_{2} \rightarrow 2CO_{2} + 3H_{2}O + SO_{2}.\]
04

Balance Sulphur Atoms

Now balance the sulfur atoms. There is one sulfur atom in both reactants and products, so they are already balanced.
05

Balance Oxygen Atoms

Lastly, balance the oxygen atoms. For the products, there are a total of 4 oxygen atoms from CO2, 3 from H2O, and 2 from SO2, summing up to 9 oxygen atoms. Therefore, 4.5 molecules of O2 are needed to balance the oxygen atoms: \[\left(\mathrm{CH}_{3}\right)_{2}S + \frac{9}{2} O_{2} \rightarrow 2CO_{2} + 3H_{2}O + SO_{2}.\] Since we cannot have a fractional coefficient in the final balanced equation, we multiply all coefficients by 2 to get whole numbers.
06

Write the Balanced Equation

Multiplying all the coefficients by 2 to eliminate the fraction gives us the final balanced equation for the combustion of dimethylsulfide: \[2\left(\mathrm{CH}_{3}\right)_{2}S + 9O_{2} \rightarrow 4CO_{2} + 6H_{2}O + 2SO_{2}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Reactions
Combustion reactions are a type of chemical reaction where a substance combines with oxygen and releases energy in the form of light or heat. Commonly, the substance that combusts is organic, containing carbon and hydrogen, and the products are predominantly carbon dioxide (CO2) and water (H2O). In the specific example of dimethylsulfide combustion, we also see the production of sulfur dioxide (SO2) due to the sulfur content in the reactant.
In a complete combustion reaction, there is enough oxygen to allow the fuel to react completely and form a limited number of products, mainly CO2 and H2O, along with any other oxides formed from other elements present in the fuel. However, if there is insufficient oxygen, incomplete combustion occurs, resulting in a more complex mix of products, often including carbon monoxide (CO) and even soot or unburned carbon particles.
Stoichiometry
Stoichiometry is the aspect of chemistry that pertains to the quantitative relationships between reactants and products in a chemical reaction. With stoichiometry, we can determine the amount of reactants needed to produce a certain amount of product, or vice versa. It's like the recipe for a chemical reaction. In our current example of the combustion of dimethylsulfide, stoichiometry helps us understand the proportions of dimethylsulfide and oxygen that react to produce carbon dioxide, water, and sulfur dioxide.

To achieve a balanced stoichiometric equation, it is crucial to ensure that the number of atoms for each element is equal on both the reactant and product sides of the equation. This adheres to the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction. Thus, the mass of the reactants must equal the mass of the products.
Chemical Reaction Balancing
Balancing chemical equations is a critical step in the study of chemistry. It ensures compliance with the Law of Conservation of Mass, indicating that atoms are neither created nor lost during a chemical reaction. The aim is to make the number of each type of atom equal on both sides of the equation. For this, we adjust the coefficients – the numbers placed before the molecules – in the reaction.

The balanced equation provides valuable insight into the stoichiometry of the reaction, allowing us to determine the relative quantities of reactants and products involved. In our exercise example, we meticulously balanced carbon, hydrogen, sulfur, and then oxygen, which required us to multiply the coefficients to get rid of any fractions and achieve whole number coefficients, which is the standard practice in representing balanced chemical equations.
Molecular Composition
The molecular composition refers to the identity and number of atoms that make up a molecule, which dictates the properties and the behavior of the molecule during chemical reactions. Understanding the molecular composition is integral in predicting product formation, explaining reactivity, and guiding the balancing of equations.
For instance, knowing that dimethylsulfide consists of two carbon (C), six hydrogen (H), and one sulfur (S) atom allows us to predict its combustion products and balance the chemical equation accordingly. As seen in our exercise, the molecular composition dictated that two molecules of carbon dioxide, three molecules of water, and one molecule of sulfur dioxide would be the products of the combustion of one molecule of dimethylsulfide in excess oxygen. This illustrates the tight link between molecular composition and stoichiometry in the systematic approach to solving chemical balance problems.

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Most popular questions from this chapter

One pollutant in smog is nitrogen dioxide, \(\mathrm{NO}_{2}\). The gas has a reddish brown color and is responsible for the redbrown color associated with this type of air pollution. \(\mathrm{Ni}\) trogen dioxide is also a contributor to acid rain because when rain passes through air contaminated with \(\mathrm{NO}_{2}\), it dissolves and undergoes the following reaction: \(\mathrm3{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)\) In this reaction, which element is reduced and which is oxidized? Which is the oxidizing agent and which is the reducing agent?

Sulfites are used worldwide in the wine industry as antioxidant and antimicrobial agents. However, sulfites have also been identified as causing certain allergic reactions suffered by asthmatics, and the FDA mandates that sulfites be identified on the label if they are present at levels of 10 ppm (parts per million) or higher. The analysis of sulfites in wine uses the "Ripper method" in which a standard iodine solution, prepared by the reaction of iodate and iodide ions, is used to titrate a sample of the wine. The iodine is formed in the reaction $$ \mathrm{IO}_{3}^{-}+5 \mathrm{I}^{-}+6 \mathrm{H}^{+} \longrightarrow 3 \mathrm{I}_{2}+3 \mathrm{H}_{2} \mathrm{O} $$ The iodine is held in solution by adding an excess of \(\mathrm{I}^{-}\), which combines with \(\mathrm{I}_{2}\) to give \(\mathrm{I}_{3}^{-}\). In the titration, the \(\mathrm{SO}_{3}^{2-}\) is converted to \(\mathrm{SO}_{2}\) by acidification, and the reaction during the titration is $$ \mathrm{SO}_{2}+\mathrm{I}_{3}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{SO}_{4}^{2-}+3 \mathrm{I}^{-}+4 \mathrm{H}^{+} $$ Starch is added to the wine sample to detect the end point, which is signaled by the formation of a dark blue color when excess iodine binds to the starch molecules. In a certain analysis, \(0.0421 \mathrm{~g}\) of \(\mathrm{NaIO}_{3}\) was dissolved in dilute acid and excess NaI was added to the solution, which was then diluted to a total volume of \(100.0 \mathrm{~mL}\) A \(50.0 \mathrm{~mL}\) sample of wine was then acidified and titrated with the iodine- containing solution. The volume of iodine solution required was \(2.47 \mathrm{~mL}\). (a) What was the molarity of the iodine (actually, \(\left.\mathrm{I}_{3}^{-}\right)\) in the standard solution? (b) How many grams of \(\mathrm{SO}_{2}\) were in the wine sample? (c) If the density of the wine was \(0.96 \mathrm{~g} / \mathrm{mL}\), what was the percentage of \(\mathrm{SO}_{2}\) in the wine? (d) Parts per million (ppm) is calculated in a manner similar to percent (which is equivalent to parts per hundred). $$ \mathrm{ppm}=\frac{\text { grams of component }}{\text { grams of sample }} \times 10^{6} \mathrm{ppm} $$ What was the concentration of sulfite in the wine, expressed as parts per million \(\mathrm{SO}_{2} ?\)

Write balanced chemical equations for the complete combustion (in the presence of excess oxygen) of the following: (a) \(\mathrm{C}_{12} \mathrm{H}_{26}\) (a component of kerosene), (b) \(\mathrm{C}_{18} \mathrm{H}_{36}\) (a component of diesel fuel), (c) \(\mathrm{C}_{7} \mathrm{H}_{8}\) (toluene, a raw material in the production of the explosive TNT).

Assign oxidation numbers to the atoms in the following: (a) \(\mathrm{ClO}_{4}^{-},\) (b) \(\mathrm{Cl}^{-}\), (c) \(\mathrm{SF}_{6}\), and (d) \(\mathrm{Au}\left(\mathrm{NO}_{3}\right)_{3}\).

For the following reactions, identify the substance oxidized, the substance reduced, the oxidizing agent, and the reducing agent. $$ \begin{array}{l} \text { (a) } 2 \mathrm{HNO}_{3}+3 \mathrm{H}_{3} \mathrm{AsO}_{3} \longrightarrow \\ 2 \mathrm{NO}+3 \mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{H}_{2} \mathrm{O} \\ \text { (b) } \mathrm{NaI}+3 \mathrm{HOCl} \longrightarrow \mathrm{NaIO}_{3}+3 \mathrm{HCl} \end{array} $$ (c) \(2 \mathrm{KMnO}_{4}+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow\) $$ 10 \mathrm{CO}_{2}+\mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+8 \mathrm{H}_{2} \mathrm{O} $$ (d) \(6 \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{Al} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{SO}_{2}+6 \mathrm{H}_{2} \mathrm{O}\)

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