Complete and balance each hydrocarbon combustion reaction. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}+\mathrm{O}_{2} \longrightarrow\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{O}_{2} \longrightarrow\) (c) \(\mathrm{CH} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{3}+\mathrm{O}_{2} \longrightarrow\)

The process of balancing a chemical reaction is critical to understanding the exact proportions in which reactants combine to form products.

For instance, when balancing hydrocarbon combustion reactions, students must ensure that the number of atoms of carbon (C), hydrogen (H), and oxygen (O) is the same on both the reactant and product sides of the equation. This ensures the law of conservation of mass is obeyed.

In a balanced combustion reaction, every carbon atom in the hydrocarbon becomes part of a carbon dioxide (CO2) molecule, and every hydrogen atom turns into part of a water (H2O) molecule. To balance the equation, you often start by balancing the carbon and hydrogen atoms first and then adjust the oxygen molecules last, as oxygen is usually present in excess from O2 in the air.

As seen in the step by step solution, the general formula for hydrocarbon combustion (\( \mathrm{C}_x\mathrm{H}_y + \frac{x + \frac{y}{4}}{2} \mathrm{O}_2 \longrightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O} \)) was used as a template to ensure the law of conservation of mass is met.

Stoichiometry is the study of the quantitative relationships or ratios between the reactants and products in a chemical reaction.

In the combustion of hydrocarbons, which are organic compounds composed solely of carbon and hydrogen, stoichiometry allows us to calculate the exact amount of oxygen needed for complete combustion and the amounts of carbon dioxide and water produced. The process is vital for many industrial applications, including engine design and environmental monitoring.

Consider the balanced chemical equation for the combustion of butane (\( \mathrm{C}_4\mathrm{H}_{10} + 13\mathrm{O}_2 \longrightarrow 8\mathrm{CO}_2 + 10\mathrm{H}_2\mathrm{O} \)). Here, stoichiometry tells us that one mole of butane reacts with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water. This ratio is crucial in ensuring the reaction goes to completion and that there is no leftover reactant.

To perform stoichiometric calculations, students must start with a correctly balanced chemical equation and use it as the basis to relate reactants to products using molar ratios.

Alkanes and alkenes are two types of hydrocarbons with different chemical properties, which largely stem from their distinct structural characteristics.

Alkanes are saturated hydrocarbons, meaning they contain only single bonds between carbon atoms and have the general formula CnH(2n+2). Examples include methane (CH4) and butane (C4H10), as in our exercise. They are known for their less reactive natures due to the strong carbon-carbon (C-C) and carbon-hydrogen (C-H) single bonds.

On the other hand, alkenes are unsaturated hydrocarbons that include at least one double bond between carbon atoms, following the general formula CnH2n. Ethene (C2H4) is an example provided in the exercise. The presence of the carbon-carbon double bond in alkenes makes them more reactive than alkanes.

When burned in air, both alkanes and alkenes undergo combustion reactions to form carbon dioxide and water. However, due to their different hydrogen-to-carbon ratios, the stoichiometry of their combustion reactions will differ, as was demonstrated in the step by step solutions for butane (an alkane) and ethene (an alkene).