Question

Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) $$ Pentene (iii) Hexyne (iv) Toluene

Short answer

Balanced combustion equations: Butane: $2C_4{H}_{10}+13O_2\to 8C{O}_2+10H_{2}O$; Pentene: $2C_5{H}_{10}+15O_2\to 10C{O}_2+10H_{2}O$; Hexyne: $2C_6{H}_{10}+17O_2\to 12C{O}_2+10H_{2}O$; Toluene: $C_7{H}_8+9O_2\to 7C{O}_2+4H_{2}O$.

Step-by-step solution

Understand Combustion Reactions

Combustion reactions involve a hydrocarbon reacting with oxygen to produce carbon dioxide and water. The general form is: $$C_x{H}_y+O_2\to C{O}_2+H_{2}O$$ Ensure balance in the number of atoms on both sides of the equation.

Write Combustion Equation for Butane

Butane is $C_4{H}_{10}$. The chemical equation for its combustion is: $$C_4{H}_{10}+O_2\to C{O}_2+H_{2}O$$. Balance the equation: $$2C_4{H}_{10}+13O_2\to 8C{O}_2+10H_{2}O$$.

Write Combustion Equation for Pentene

Pentene is $C_5{H}_{10}$. Write its combustion equation: $$C_5{H}_{10}+O_2\to C{O}_2+H_{2}O$$. Balance it: $$2C_5{H}_{10}+15O_2\to 10C{O}_2+10H_{2}O$$.

Write Combustion Equation for Hexyne

Hexyne is $C_6{H}_{10}$. Write its combustion equation: $$C_6{H}_{10}+O_2\to C{O}_2+H_{2}O$$. Balance it: $$2C_6{H}_{10}+17O_2\to 12C{O}_2+10H_{2}O$$.

Write Combustion Equation for Toluene

Toluene is $C_7{H}_8$. Write its combustion equation: $$C_7{H}_8+O_2\to C{O}_2+H_{2}O$$. Balance it: $$C_7{H}_8+9O_2\to 7C{O}_2+4H_{2}O$$.

Key concepts

Understanding Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They illustrate the substances involved, from reactants to products. On one side of the equation, you have the reactants, the substances involved in the reaction's starting point.
On the other side, you have the products, the new substances formed from the reaction. Chemical equations are like equations in math; they need to be balanced to accurately represent the conservation of mass. Balancing is crucial because it ensures that the number of atoms of each element is the same on both sides of the equation.
This balance reflects the principle of conservation of matter, meaning atoms are not created or destroyed in a chemical reaction. The process of balancing a chemical equation involves adjusting the coefficients (numbers in front of molecules) without changing the subscripts (numbers within molecules). For example, in a typical combustion reaction where a hydrocarbon reacts with oxygen, you might start with: $$C_x{H}_y+O_2\to C{O}_2+H_{2}O$$ Balancing such an equation involves ensuring that the number of carbon, hydrogen, and oxygen atoms is identical on both sides of the equation. This might require multiplying each molecule by certain coefficients until you achieve equilibrium between reactants and products.

Hydrocarbons in Combustion Reactions

Hydrocarbons are organic compounds consisting solely of carbon and hydrogen atoms. They can be classified into different types: alkanes (single bonds), alkenes (at least one double bond), and alkynes (at least one triple bond).
These structures affect the number of hydrogen atoms in the molecule and consequently influence the balancing of combustion equations. In a combustion reaction, hydrocarbons are burned in the presence of oxygen, resulting in complete combustion to carbon dioxide and water.
Let's break it down with examples: - **Butane ( C_4H_{10} ):** As an alkane, it combusts to form carbon dioxide and water, making it relatively straightforward to balance. - **Pentene ( C_5H_{10} ):** An alkene, which means it has less hydrogen per carbon due to the double bond, requiring careful balancing. - **Hexyne ( C_6H_{10} ):** Being an alkyne, it features a triple bond, demanding even more precision when balancing due to its lower hydrogen content. - **Toluene ( C_7H_8 ):** A derivative of benzene, forming carbon dioxide and water, plus needing precise balancing to manage the hydrogen to carbon ratio. Each hydrocarbon type provides a unique balancing challenge depending on its molecular structure, but the end result in combustion is always production of carbon dioxide and water if combustion is complete.

Stoichiometry and Its Role in Combustion

Stoichiometry is the calculation method used to find relative quantities of reactants and products in chemical reactions. It's essential in balancing chemical equations, especially for reactions like combustion, where precise ratios dictate whether the equation is balanced. In stoichiometry, molar ratios derived from balanced equations are employed to calculate the amount of reactions or products. This involves using coefficients as conversion factors to move from moles of one substance to moles of another.
Let's explore how stoichiometry is applied in combustion reactions: - We consider the **mole-to-mole relationships**, which tell you proportions between reactants and products. This helps in predicting the amounts of products formed. - **For Butane:** The combustion equation involves multiplying butane by 2. This adjusts the equation so that it corresponds to the 8 moles of carbon dioxide and 10 moles of water on the products side. - **For Toluene:** Here, stoichiometric calculations help in accounting for its 7 carbon atoms, aligning them to the formation of 7 moles of carbon dioxide while managing oxygen and hydrogen balance. By engaging stoichiometry, one can balance the equation perfectly, providing a quantitative basis for determining how much of each substance is involved, ensuring the reaction depicts real-life chemical transformations accurately.
As a toolbox in chemistry, stoichiometry enables the prediction and measurement of outcomes in chemical processes effectively.