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Complete and balance the following chemical equations for the complete combustion of several hydrocarbons. a. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow\) b. \(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow\) c. \(C_{6} H_{6}(\ell)+O_{2}(g) \rightarrow\) d. \(\mathrm{C}_{8} \mathrm{H}_{18}(\ell)+\mathrm{O}_{2}(g) \rightarrow\)

Short Answer

Expert verified
\) b. \(C_{4}H_{10}(g) + O_{2}(g) \rightarrow ?\) c. \(C_{6}H_{6}(l) + O_{2}(g) \rightarrow ?\) d. \(C_{8}H_{18}(l) + O_{2}(g) \rightarrow ?\) Answer: a. \(C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(l)\) b. \(C_{4}H_{10}(g) + \frac{13}{2}O_{2}(g) \rightarrow 4CO_{2}(g) + 5H_{2}O(l)\) c. \(C_{6}H_{6}(l) + \frac{15}{2}O_{2}(g) \rightarrow 6CO_{2}(g) + 3H_{2}O(l)\) d. \(C_{8}H_{18}(l) + \frac{25}{2}O_{2}(g) \rightarrow 8CO_{2}(g) + 9H_{2}O(l)\)

Step by step solution

01

Write down the products of combustion

The complete combustion of hydrocarbons always produces carbon dioxide (CO2) and water (H2O). So, the unbalanced equation will look like this: \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l)\)
02

Balance the chemical equation

To balance the equation, first balance the carbon atoms, then hydrogen atoms, and finally, the oxygen atoms. \(1\mathrm{C}_{3} \mathrm{H}_{8}(g) + 5\mathrm{O}_{2}(g) \rightarrow 3\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow\)
03

Write down the products of combustion

The complete combustion of hydrocarbons always produces carbon dioxide (CO2) and water (H2O). So, the unbalanced equation will look like this: \(\mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l)\)
04

Balance the chemical equation

To balance the equation, first balance the carbon atoms, then hydrogen atoms, and finally, the oxygen atoms. \(1\mathrm{C}_{4} \mathrm{H}_{10}(g) + \frac{13}{2}\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l)\) c. \(C_{6} H_{6}(\ell)+O_{2}(g) \rightarrow\)
05

Write down the products of combustion

The complete combustion of hydrocarbons always produces carbon dioxide (CO2) and water (H2O). So, the unbalanced equation will look like this: \(C_{6} H_{6}(\ell)+O_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l)\)
06

Balance the chemical equation

To balance the equation, first balance the carbon atoms, then hydrogen atoms, and finally, the oxygen atoms. \(1\mathrm{C}_{6} \mathrm{H}_{6}(\ell) + \frac{15}{2}\mathrm{O}_{2}(g) \rightarrow 6\mathrm{CO}_{2}(g) + 3\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{C}_{8} \mathrm{H}_{18}(\ell)+\mathrm{O}_{2}(g) \rightarrow\)
07

Write down the products of combustion

The complete combustion of hydrocarbons always produces carbon dioxide (CO2) and water (H2O). So, the unbalanced equation will look like this: \(\mathrm{C}_{8} \mathrm{H}_{18}(\ell)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l)\)
08

Balance the chemical equation

To balance the equation, first balance the carbon atoms, then hydrogen atoms, and finally, the oxygen atoms. \(1\mathrm{C}_{8} \mathrm{H}_{18}(\ell) + \frac{25}{2}\mathrm{O}_{2}(g) \rightarrow 8\mathrm{CO}_{2}(g) + 9\mathrm{H}_{2} \mathrm{O}(l)\)

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Key Concepts

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

Hydrocarbons Combustion
Combustion is a chemical process that occurs when a substance reacts quickly with oxygen and releases energy in the form of light and heat. Hydrocarbons, which are compounds made up of hydrogen and carbon, are commonly used as fuels. During combustion, these hydrocarbons combine with oxygen to produce carbon dioxide (CO鈧) and water (H鈧侽). This type of reaction is known as complete combustion.
Complete combustion is important because it minimizes the production of pollutants such as carbon monoxide (CO), which occurs during incomplete combustion when there is insufficient oxygen. To facilitate complete combustion, oxygen needs to be present in the right amount. This ensures that carbon atoms in the hydrocarbon are fully oxidized to CO鈧, and hydrogen atoms are converted to H鈧侽.
For instance, when propane (C鈧僅鈧) undergoes combustion, it reacts with oxygen to form carbon dioxide and water. This can be represented as follows:
\[\mathrm{C}_{3}\mathrm{H}_{8}(g) + 5\mathrm{O}_{2}(g) \rightarrow 3\mathrm{CO}_{2}(g) + 4\mathrm{H}_{2} \mathrm{O}(l)\]
Here, the balanced equation shows that one molecule of propane requires five molecules of oxygen to produce three molecules of carbon dioxide and four molecules of water.
Chemical Reactions
Chemical reactions are processes where reactants are transformed into products through the making and breaking of chemical bonds. In the context of combustion, hydrocarbons react with oxygen in a highly exothermic reaction, which means that it releases energy. This type of chemical reaction is crucial for many applications, from heating homes to powering vehicles.
All reactions, including combustion, follow the law of conservation of mass, which states that matter is not created or destroyed in a chemical reaction. This is why balancing the chemical equation is important鈥攖o ensure that the number of atoms for each element is the same on both sides of the equation.
Each type of chemical reaction has its unique characteristics. For combustion:
  • Oxygen is always a reactant.
  • Heat and often light are released.
  • Produces oxides, typically carbon dioxide and water in the case of hydrocarbon fuels.
Balancing the chemical equation for a combustion reaction involves making sure that the number of carbon and hydrogen atoms from the fuel matches the number of carbon dioxide and water molecules produced.
Stoichiometry
Stoichiometry is the area of chemistry that revolves around the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced. This is especially useful in industrial and laboratory settings where precise amounts of chemicals are needed.
In balancing chemical equations, stoichiometry is used to ensure that the reactants and products adhere to the conservation of mass. Consider the combustion of butane (C鈧凥鈧佲個):
\[\mathrm{C}_{4} \mathrm{H}_{10}(g) + \frac{13}{2}\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l)\]
The balanced equation shows exactly how many moles of oxygen are needed to completely react with one mole of butane to yield carbon dioxide and water. Stoichiometry enables you to scale up this reaction for practical uses by multiplying the balanced equation by whatever factor necessary to get the desired amount of product.
In summary, stoichiometry provides the tools to balance chemical equations and understand the proportions of elements that combine in chemical reactions, ensuring efficient use of materials and energy.

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Most popular questions from this chapter

Some indoor air-purification systems work by converting a little of the oxygen in the air to ozone, which kills mold and mildew spores and other biological air pollutants. The chemical equation for the ozone generation reaction is $$3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{O}_{3}(g)$$ It is claimed that one such system generates \(4.0 \mathrm{g} \mathrm{O}_{3}\) per hour from dry air passing through the purifier at a flow of \(5.0 \mathrm{L} / \mathrm{min.}\) If exactly 1 liter of indoor air contains \(0.28 \mathrm{g}\) \(\mathrm{O}_{2},\) what percentage of the \(\mathrm{O}_{2}\) is converted to \(\mathrm{O}_{3}\) by the air purifier?

Black Powder Gunpowder, known also as black powder, is generally thought to have been invented by the Chinese in the 9 th century. The first description in English was given by Roger Bacon in the 13 th century. The basic formulation has not changed since then: \(40 \%\) potassium nitrate, \(30 \%\) carbon, and \(30 \%\) sulfur by weight. The products of the reaction that provide the explosive force when the powder is ignited are three gases: carbon monoxide, carbon dioxide, and nitrogen. An additional product is potassium sulfite. Consider a sample of \(100 \mathrm{g}\) of black powder. a. Which reagent is limiting? b. How much of each of the reagents in excess is left over after the explosion takes place?

Some catalytic converters in automobiles contain two manganese oxides: \(\mathrm{Mn}_{2} \mathrm{O}_{3}\) and \(\mathrm{MnO}_{2}.\) a. What are the names of these compounds? b. What is the manganese content of each (expressed as a percent by mass)? c. Explain how \(\mathrm{Mn}_{2} \mathrm{O}_{3}\) and \(\mathrm{MnO}_{2}\) are consistent with the law of multiple proportions.

Why is the quantity of \(\mathrm{CO}_{2}\) obtained in a combustion analysis not a direct measure of the oxygen content of the starting compound?

Pigments for Stoplights Cadmium yellow (cadmium sulfide) is a lemon-yellow pigment used in the lenses of stoplights. Its formula is CdS, and it is very insoluble in water. The recommended recipe for cadmium yellow is to mix cadmium nitrate with sodium sulfide in water. The cadmium yellow forms as a solid, while the other product, sodium nitrate, remains dissolved in the water. a. Write a balanced chemical equation for the reaction. b. Calculate the mass of cadmium nitrate you must start with to make 125 g of CdS.

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