Problem 7 - Chemistry in 30 days

Question

About \(15 \mathrm{~mL}\) of a gaseous hydrocarbon requires \(45 \mathrm{~mL}\) of oxygen for complete combustion, which produces \(30 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) gas. The formula of the hydrocarbon is (a) \(\mathrm{C}_{3} \mathrm{H}_{6}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{4}\)

Step-by-Step Solution

Answer

The hydrocarbon is \( C_2H_6 \). Option (b).

Step 1: Write the General Reaction for Combustion

A hydrocarbon, represented as \( C_xH_y \), undergoes combustion with oxygen to produce carbon dioxide and water vapor:\[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \] Our task is to determine the values of \( x \) and \( y \) that will satisfy the volumes given in the problem.

Step 2: Interpret the Given Data

The problem statement provides the following volumetric data for the complete combustion:- 15 mL of hydrocarbon- 45 mL of oxygen required- 30 mL of \( CO_2 \) producedUsing the molar volume concept, these volumes represent the stoichiometric coefficients in the equation because volumes of gases at the same temperature and pressure are directly proportional to the number of moles.

Step 3: Write the Balanced Equation for Combustion

For a general hydrocarbon \( C_xH_y \), the balanced equation for combustion in terms of moles is:\[ C_xH_y + \frac{3x + y}{4} O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O \] From the data:- The volume of \( CO_2 \) is 30 mL which implies \( x = 2 \) (to produce 30 mL of \( CO_2 \) from 15 mL of hydrocarbon).- The stoichiometry for oxygen utilized is 45 mL, which should match with \( \frac{3x + y}{4} = 45/15 = 3 \) as moles ratio.

Step 4: Solve for \( x \) and \( y \)

Given \( x = 2 \) from the \( CO_2 \) volume, substitute in:\[ \frac{3(2) + y}{4} = 3 \]Solve for \( y \):\[ \frac{6 + y}{4} = 3 \]\[ 6 + y = 12 \]\[ y = 6 \]

Step 5: Confirm the Hydrocarbon Formula

Using the values \( x = 2 \) and \( y = 6 \), the formula for the hydrocarbon is determined to be \( C_2H_6 \).

Step 6: Compare with Options

Compare the derived formula \( C_2H_6 \) with the provided options:(a) \( C_3H_6 \) (b) \( C_2H_6 \) (c) \( C_4H_{10} \) (d) \( C_2H_4 \)The correct formula is option (b) \( C_2H_6 \).

Key Concepts

Chemical Equation Balancing

Balancing a chemical equation is essential for representing the conservation of mass in a chemical reaction. It ensures the number of atoms of each element is the same on both sides of the equation. In the combustion of hydrocarbons, we start with a generic formula for the hydrocarbon as \( C_xH_y \). During complete combustion, this reacts with oxygen (\( O_2 \)) to produce carbon dioxide (\( CO_2 \)) and water (\( H_2O \)). The general balanced equation can be written as:

• \[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

The coefficients of the equation are adjusted to ensure that the same number of each type of atom appears on both sides. The balancing process is somewhat trial and error, where numbers are tweaked to achieve balance. This ensures that the law of conservation of mass is followed and is critical for accurate chemical analysis.

Molar Volume in Gases

The concept of molar volume is crucial for understanding gas reactions. At standard temperature and pressure (STP), one mole of any gas occupies a volume of 22.4 liters. This principle allows us to relate the volume of gases to moles, which is the basis for stoichiometry in gaseous reactions. In the given problem, we used the ratios of the volumes of gases directly because they represent the stoichiometric coefficients in the balanced chemical equation. For example, 15 mL of hydrocarbon requiring 45 mL of \( O_2 \) and producing 30 mL of \( CO_2 \) helps deduce the mole ratio:

• Hydrocarbon : Oxygen : Carbon Dioxide = 15 : 45 : 30 = 1 : 3 : 2

This proportionality allows for determining the coefficients in the balanced equation based on experimental data. This concept simplifies the analysis of combustion and many other gas-related reactions.

Stoichiometry in Chemistry

Stoichiometry is the calculation of reactants and products in a chemical reaction. It is an integral part of chemistry that uses balanced equations to determine the amounts of substances consumed and produced. In a combustion reaction scenario, stoichiometry helps map out the relationship between the reactant volumes and product volumes. Using the balanced equation:

• \[ C_xH_y + \frac{3x + y}{4} O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O \]

The coefficients provide the ratio of moles for each compound. By interpreting the chemical equation in this way, it's possible to solve for unknowns, like the molecular formula of the hydrocarbon. Thus, stoichiometry allows us to translate theoretical and experimental data into meaningful chemical formulas and quantities.

Combustion Reaction

Combustion is a type of exothermic reaction where a substance reacts with oxygen to release energy in the form of heat and light. In the context of hydrocarbon combustion, it involves burning a hydrocarbon in sufficient oxygen, producing carbon dioxide and water as byproducts. This is expressed in the chemical equation:

• \[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

Key features include:

• The need for oxygen as a reactant.
• The production of \( CO_2 \) and \( H_2O \).
• The release of energy, often making combustion reactions self-sustaining once initiated.

The completeness of a combustion reaction depends on having sufficient oxygen to convert all the carbon in the hydrocarbon to carbon dioxide, rather than partial oxidation products. This reaction is fundamental not only in chemistry but also in various industrial processes and everyday life applications.